{"id":16178,"date":"2017-06-27T16:56:50","date_gmt":"2017-06-27T16:56:50","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/working-with-logarithms\/"},"modified":"2017-06-27T16:56:50","modified_gmt":"2017-06-27T16:56:50","slug":"working-with-logarithms","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/working-with-logarithms\/","title":{"raw":"Working With Logarithms","rendered":"Working With Logarithms"},"content":{"raw":"

Logarithms of Products<\/h2>\nA useful property of logarithms states that the logarithm of a product of two quantities is the sum of the logarithms of the two factors. In symbols,\u00a0[latex]\\log_b(xy)=\\log_b(x)+\\log_b(y).[\/latex]\n
\n

Learning Objectives<\/h3>\nRelate the product rule for logarithms to the rules for operating with exponents, and use this rule to rewrite logarithms of products\n\n<\/div>\n
\n

Key Takeaways<\/h3>\n

Key Points<\/h4>\n
    \n \t
  • The logarithm of a product is the sum of the logarithms of the factors.<\/li>\n \t
  • The product rule does not apply when the base of the two logarithms are different.<\/li>\n<\/ul>\n

    Key Terms<\/h4>\n
      \n \t
    • exponent<\/strong>: The power to which a number, symbol, or expression is to be raised. For example, the 3 in [latex]x^3[\/latex].<\/li>\n<\/ul>\n<\/div>\n

      Logarithms<\/h3>\nThe logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number. For example, the logarithm of [latex]1000[\/latex] in base [latex]10[\/latex] is [latex]3[\/latex], because [latex]10^3=1000.[\/latex]\n\nMore generally, if [latex]x=b^y[\/latex], then [latex]y[\/latex] is the logarithm base [latex]b[\/latex] of [latex]x[\/latex], written: [latex]y=\\log_b(x)[\/latex], so [latex]\\log_{10}(1000)=3[\/latex].\n\nIt is useful to think of logarithms as inverses of exponentials. So, for example:\n\n[latex]\\displaystyle\n\\log_b(b^z)=z[\/latex]\n\nAnd:\n\n[latex]\\displaystyle\nb^{\\log_b(z)}=z[\/latex]\n

      Product Rule for Logarithms<\/h3>\nLogarithms were introduced by John Napier in the early 17th century as a means to simplify calculations. Logarithms were rapidly adopted by navigators, scientists, engineers, and others to perform computations more easily by using slide rules and logarithm tables. Tedious multi-digit multiplication steps can be replaced by table look-ups and simpler addition, because of the fact that the logarithm of a product is the sum of the logarithms of the factors:\n\n[latex]\\displaystyle\nlog_b(xy) = log_b(x) + log_b(y)[\/latex]\n\nWe can see that this rule is true by writing the logarithms in terms of exponentials.\n\nLet [latex]\\log_b(x)=v[\/latex]\u00a0and [latex]\\log_b(y)=w.[\/latex]\n\nWriting these equations as exponentials:\n\n[latex]\\displaystyle\nb^v=x[\/latex]\n\nAnd:\n\n[latex]\\displaystyle\nb^w=y.[\/latex]\n\nThen note that:\n\n[latex]\\displaystyle\n\\begin{align}\nxy&=b^vb^w\\\\\n&=b^{v+w}\n\\end{align}[\/latex]\n\nTaking the logarithm base [latex]b[\/latex]\u00a0of both sides of this last equation yields:\n\n[latex]\\displaystyle\n\\begin{align}\n\\log_b(xy)&=\\log_b(b^{v+w})\\\\\n&=v+w\\\\\n&=\\log_b(x) + \\log_b(y)\n\\end{align}[\/latex]\n\nThis is a very useful property of logarithms, because it can sometimes simplify more complex expressions. For example:\n\n[latex]\\displaystyle\n\\log_{10}(10^x\\cdot 100^{x^3+1})=\\log_{10} (10^x)+\\log_{10}(100^{x^3+1}) [\/latex]\n\nThen because [latex]100[\/latex]\u00a0is [latex]10^2[\/latex], we have:\n\n[latex]\\displaystyle\n\\begin{align}\nx+\\log_{10}(10^{2(x^3+1)}) &= x+2(x^3+1)\\\\\n&=2x^3+x+2\n\\end{align}[\/latex]\n

      Logarithms of Powers<\/h2>\nThe logarithm of the [latex]p\\text{th}[\/latex] power of a quantity is [latex]p[\/latex] times the logarithm of the quantity. In symbols,\u00a0[latex]\\log_b(x^p)=p\\log_b(x).[\/latex]\n
      \n

      Learning Objectives<\/h3>\nRelate the power rule for logarithms to the rules for operating with exponents, and use this rule to rewrite logarithms of powers\n\n<\/div>\n
      \n

      Key Takeaways<\/h3>\n

      Key Points<\/h4>\n
        \n \t
      • The logarithm of a product is the sum of the logarithms of the factors.<\/li>\n \t
      • An exponent, [latex]p[\/latex], signifies that a number is being multiplied by itself [latex]p[\/latex] number of times. Because the logarithm of a product is the sum of the logarithms of the factors, the logarithm of a number, [latex]x[\/latex], to an exponent, [latex]p[\/latex], is the same as the logarithm of [latex]x[\/latex]\u00a0added together [latex]p[\/latex] times, so it is equal to\u00a0[latex]p\\log_b(x).[\/latex]<\/li>\n<\/ul>\n

        Key Terms<\/h4>\n
          \n \t
        • base<\/strong>: A number raised to the power of an exponent.<\/li>\n \t
        • logarithm<\/strong>: The logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number.<\/li>\n \t
        • exponent<\/strong>: The power to which a number, symbol, or expression is to be raised. For example, the 3 in [latex]x^3[\/latex].<\/li>\n<\/ul>\n<\/div>\n

          The Power Rule for Logarithms<\/h3>\nWe have already seen that the logarithm of a product is the sum of the logarithms of the factors:\n\n[latex]\\displaystyle\n\\log _b \\left( {xy} \\right) = \\log _b \\left( x \\right) + \\log _b \\left( y \\right)[\/latex]\n\nIf we apply this rule repeatedly we can devise another rule for simplifying expressions of the form [latex]\\log_b x^p[\/latex].\n\nRecall that [latex]x^p[\/latex]\u00a0can be thought of as [latex]x \\cdot x \\cdot x \\cdots x[\/latex]\u00a0where there are [latex]p[\/latex]\u00a0factors of [latex]x[\/latex]. Then we have:\n\n[latex]\\displaystyle\n\\begin{align}\n\\log_b(x^p) &= \\log_b (x \\cdot x \\cdots x) \\\\\n&= \\log_b x + \\log_b x + \\cdots +\\log_b x \\\\\n&= p\\log_b x\n\\end{align}[\/latex]\n\nSince the [latex]p[\/latex]\u00a0factors of [latex]x[\/latex]\u00a0are converted to [latex]p [\/latex]\u00a0summands by the product rule formula.\n

          Example 1: Simplify the expression\u00a0[latex]\\log_3(3^x\\cdot 9x^{100})[\/latex]<\/h3>\nFirst expand the log:\n\n[latex]\\displaystyle\n\\log_3(3^x\\cdot 9x^{100}) =\\log_3 (3^x) + \\log_3 9 + \\log_3(x^{100}) [\/latex]\n\nNext use the product and power rule to simplify:\n\n[latex]\\displaystyle\n\\log_3 (3^x) + \\log_3 9 + \\log_3 (x^{100})= x+2+100\\log_3 x[\/latex]\n

          Example 2: Solve [latex]2^{(x+1)}=10^3[\/latex]\u00a0for [latex]x[\/latex]\u00a0using logarithms<\/h3>\nStart\u00a0by taking the logarithm with base [latex]2[\/latex]\u00a0of both sides:\n\n[latex]\\displaystyle\n\\begin{align}\n\\log_2 (2^{(x+1)}) &= \\log_2 (10^3)\\\\\nx+1&=3\\log_2(10)\\\\\nx&=3\\log_2(10)-1\n\\end{align}[\/latex]\n\nTherefore a solution would be\u00a0[latex]x=3\\log_2(10) -1. [\/latex]\n

          Logarithms of Quotients<\/h2>\nThe logarithm of the ratio of two quantities is the difference of the logarithms of the quantities. In symbols, [latex]\\log_b\\left( \\frac{x}{y}\\right) = \\log_bx - \\log_by.[\/latex]\n
          \n

          Learning Objectives<\/h3>\nRelate the quotient rule for logarithms to the rules for operating with exponents, and use this rule to rewrite logarithms of quotients\n\n<\/div>\n
          \n

          Key Takeaways<\/h3>\n

          Key Points<\/h4>\n
            \n \t
          • The logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number.<\/li>\n \t
          • The logarithm of a product is the sum of the logarithms of the factors.<\/li>\n \t
          • The logarithm of the ratio or quotient of two numbers is the difference of the logarithms.<\/li>\n<\/ul>\n

            Key Terms<\/h4>\n
              \n \t
            • exponent<\/strong>: The power to which a number, symbol, or expression is to be raised. For example, the 3 in [latex]x^3[\/latex].<\/li>\n<\/ul>\n<\/div>\nWe have already seen that the logarithm of a product is the sum of the logarithms of the factors:\n\n[latex]\\displaystyle\n\\log_b(xy) = \\log_bx + \\log_by[\/latex]\n\nSimilarly, the logarithm of the ratio of two quantities is the difference of the logarithms:\n\n[latex]\\displaystyle\n\\log_b\\left( \\frac{x}{y}\\right) = \\log_bx -\nlog_by.[\/latex]\n\nWe can show that this is true by the following example:\n\nLet [latex]u=\\log_b x[\/latex]\u00a0and [latex]v=\\log_b y[\/latex].\n\nThen [latex]b^u=x[\/latex]\u00a0and [latex]b^v=y.[\/latex]\n\nThen:\n\n[latex]\\displaystyle\n\\begin{align}\n\\log_b\\left(\\frac{x}{y}\\right)&=\\log_b\\left({b^u \\over b^v}\\right)\\\\\n&= \\log_b(b^{u-v}) \\\\\n&=u-v\\\\\n&= \\log_b x - \\log_b y\n\\end{align}[\/latex]\n\nAnother way to show that this rule is true, is to apply both the power and product rules and the fact that dividing by [latex]y[\/latex]\u00a0is the same is multiplying by [latex]y^{-1}.[\/latex]\u00a0So we can write:\n\n[latex]\\displaystyle\n\\begin{align}\n\\log_b\\left(\\frac{x}{y}\\right)&=\\log_b(x\\cdot y^{-1})\\\\\n& = \\log_bx + \\log_b(y^{-1})\\\\&\n= \\log_bx -\\log_by\n\\end{align}[\/latex]\n

              Example: write the expression [latex]\\log_2\\left({x^4y^9 \\over z^{100}}\\right)[\/latex]\u00a0in a simpler way<\/h3>\nBy applying the product, power, and quotient rules, you could write this expression as:\n\n[latex]\\displaystyle\n\\log_2(x^4)+\\log_2(y^9)-\\log_2(z^{100}) = 4\\log_2x+9\\log_2y-100\\log_2z.[\/latex]\n

              Changing Logarithmic Bases<\/h2>\nA logarithm written in one base can be converted to an equal quantity written in a different base.\n
              \n

              Learning Objectives<\/h3>\nUse the change of base formula to convert logarithms to different bases\n\n<\/div>\n
              \n

              Key Takeaways<\/h3>\n

              Key Points<\/h4>\n
                \n \t
              • The base of a logarithm can be changed by expressing it as the quotient of two logarithms with a common\u00a0base.<\/li>\n \t
              • Changing a logarithm's base to [latex]10[\/latex] makes it much simpler to evaluate; it can be done on a calculator.<\/li>\n<\/ul>\n

                Key Terms<\/h4>\n
                  \n \t
                • logarithm<\/strong>: The logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number.<\/li>\n \t
                • base<\/strong>: A number raised to the power of an exponent.<\/li>\n<\/ul>\n<\/div>\nMost common scientific calculators have a key for computing logarithms with base [latex]10[\/latex], but do not have keys for other bases. So, if you needed to get an approximation to a number like [latex]\\log_4(9)[\/latex]\u00a0it can be difficult to do so. One could easily guess that it is between [latex]1[\/latex]\u00a0and [latex]2[\/latex]\u00a0since [latex]9[\/latex]\u00a0is between [latex]4^1[\/latex]\u00a0and [latex]4^2[\/latex], but it is difficult to get an accurate approximation. Fortunately, there is a change of base formula that can help.\n

                  Change of Base Formula<\/h3>\nThe change of base formula for logarithms is:\n\n[latex]\\displaystyle\n\\log_a(x)=\\frac{\\log_b(x)}{\\log_b(a)}[\/latex]\n\nThus, for example, we could calculate that [latex]\\log_4(9)=\\frac{\\log_{10}(9)}{\\log_{10}(4)}[\/latex]\u00a0which could be computed on almost any handheld calculator.\n

                  Deriving the Formula<\/h3>\nTo see why the formula is true, give [latex]\\log_a(x)[\/latex]\u00a0a name like [latex]z[\/latex]:\n\n[latex]\\displaystyle\nz=\\log_a(x)[\/latex]\n\nWrite this as\u00a0[latex]a^z=x[\/latex]\n\nNow take the logarithm with base [latex]b[\/latex]\u00a0of both sides, yielding:\n\n[latex]\\displaystyle\n\\log_b a^z = \\log_bx[\/latex]\n\nUsing the power rule gives:\n\n[latex]\\displaystyle\nz \\cdot \\log_ba = \\log_b x[\/latex]\n\nDividing both sides by [latex]\\log_ba[\/latex]\u00a0gives:\n\n[latex]\\displaystyle\nz={\\log_b x \\over \\log_ba}.[\/latex]\n\nThus we have\u00a0[latex]\\log_a x ={\\log_b x \\over \\log_b a}. [\/latex]\n

                  Example<\/h3>\nAn expression of the form [latex]\\log_5(10^{x^2+1})[\/latex]\u00a0might be easier to graph on a graphing calculator or other device if it were written in base [latex]10[\/latex]\u00a0instead of base 5. The change-of-base formula can be applied to it:\n\n[latex]\\displaystyle\n\\log_5(10^{x^2+1}) = {\\log_{10}(10^{x^2+1}) \\over \\log_{10}5}[\/latex]\n\nWhich can be written as\u00a0[latex]{x^2+1 \\over \\log_{10} 5}. [\/latex]","rendered":"

                  Logarithms of Products<\/h2>\n

                  A useful property of logarithms states that the logarithm of a product of two quantities is the sum of the logarithms of the two factors. In symbols,\u00a0[latex]\\log_b(xy)=\\log_b(x)+\\log_b(y).[\/latex]<\/p>\n

                  \n

                  Learning Objectives<\/h3>\n

                  Relate the product rule for logarithms to the rules for operating with exponents, and use this rule to rewrite logarithms of products<\/p>\n<\/div>\n

                  \n

                  Key Takeaways<\/h3>\n

                  Key Points<\/h4>\n
                    \n
                  • The logarithm of a product is the sum of the logarithms of the factors.<\/li>\n
                  • The product rule does not apply when the base of the two logarithms are different.<\/li>\n<\/ul>\n

                    Key Terms<\/h4>\n
                      \n
                    • exponent<\/strong>: The power to which a number, symbol, or expression is to be raised. For example, the 3 in [latex]x^3[\/latex].<\/li>\n<\/ul>\n<\/div>\n

                      Logarithms<\/h3>\n

                      The logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number. For example, the logarithm of [latex]1000[\/latex] in base [latex]10[\/latex] is [latex]3[\/latex], because [latex]10^3=1000.[\/latex]<\/p>\n

                      More generally, if [latex]x=b^y[\/latex], then [latex]y[\/latex] is the logarithm base [latex]b[\/latex] of [latex]x[\/latex], written: [latex]y=\\log_b(x)[\/latex], so [latex]\\log_{10}(1000)=3[\/latex].<\/p>\n

                      It is useful to think of logarithms as inverses of exponentials. So, for example:<\/p>\n

                      [latex]\\displaystyle \\log_b(b^z)=z[\/latex]<\/p>\n

                      And:<\/p>\n

                      [latex]\\displaystyle b^{\\log_b(z)}=z[\/latex]<\/p>\n

                      Product Rule for Logarithms<\/h3>\n

                      Logarithms were introduced by John Napier in the early 17th century as a means to simplify calculations. Logarithms were rapidly adopted by navigators, scientists, engineers, and others to perform computations more easily by using slide rules and logarithm tables. Tedious multi-digit multiplication steps can be replaced by table look-ups and simpler addition, because of the fact that the logarithm of a product is the sum of the logarithms of the factors:<\/p>\n

                      [latex]\\displaystyle log_b(xy) = log_b(x) + log_b(y)[\/latex]<\/p>\n

                      We can see that this rule is true by writing the logarithms in terms of exponentials.<\/p>\n

                      Let [latex]\\log_b(x)=v[\/latex]\u00a0and [latex]\\log_b(y)=w.[\/latex]<\/p>\n

                      Writing these equations as exponentials:<\/p>\n

                      [latex]\\displaystyle b^v=x[\/latex]<\/p>\n

                      And:<\/p>\n

                      [latex]\\displaystyle b^w=y.[\/latex]<\/p>\n

                      Then note that:<\/p>\n

                      [latex]\\displaystyle \\begin{align} xy&=b^vb^w\\\\ &=b^{v+w} \\end{align}[\/latex]<\/p>\n

                      Taking the logarithm base [latex]b[\/latex]\u00a0of both sides of this last equation yields:<\/p>\n

                      [latex]\\displaystyle \\begin{align} \\log_b(xy)&=\\log_b(b^{v+w})\\\\ &=v+w\\\\ &=\\log_b(x) + \\log_b(y) \\end{align}[\/latex]<\/p>\n

                      This is a very useful property of logarithms, because it can sometimes simplify more complex expressions. For example:<\/p>\n

                      [latex]\\displaystyle \\log_{10}(10^x\\cdot 100^{x^3+1})=\\log_{10} (10^x)+\\log_{10}(100^{x^3+1})[\/latex]<\/p>\n

                      Then because [latex]100[\/latex]\u00a0is [latex]10^2[\/latex], we have:<\/p>\n

                      [latex]\\displaystyle \\begin{align} x+\\log_{10}(10^{2(x^3+1)}) &= x+2(x^3+1)\\\\ &=2x^3+x+2 \\end{align}[\/latex]<\/p>\n

                      Logarithms of Powers<\/h2>\n

                      The logarithm of the [latex]p\\text{th}[\/latex] power of a quantity is [latex]p[\/latex] times the logarithm of the quantity. In symbols,\u00a0[latex]\\log_b(x^p)=p\\log_b(x).[\/latex]<\/p>\n

                      \n

                      Learning Objectives<\/h3>\n

                      Relate the power rule for logarithms to the rules for operating with exponents, and use this rule to rewrite logarithms of powers<\/p>\n<\/div>\n

                      \n

                      Key Takeaways<\/h3>\n

                      Key Points<\/h4>\n
                        \n
                      • The logarithm of a product is the sum of the logarithms of the factors.<\/li>\n
                      • An exponent, [latex]p[\/latex], signifies that a number is being multiplied by itself [latex]p[\/latex] number of times. Because the logarithm of a product is the sum of the logarithms of the factors, the logarithm of a number, [latex]x[\/latex], to an exponent, [latex]p[\/latex], is the same as the logarithm of [latex]x[\/latex]\u00a0added together [latex]p[\/latex] times, so it is equal to\u00a0[latex]p\\log_b(x).[\/latex]<\/li>\n<\/ul>\n

                        Key Terms<\/h4>\n
                          \n
                        • base<\/strong>: A number raised to the power of an exponent.<\/li>\n
                        • logarithm<\/strong>: The logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number.<\/li>\n
                        • exponent<\/strong>: The power to which a number, symbol, or expression is to be raised. For example, the 3 in [latex]x^3[\/latex].<\/li>\n<\/ul>\n<\/div>\n

                          The Power Rule for Logarithms<\/h3>\n

                          We have already seen that the logarithm of a product is the sum of the logarithms of the factors:<\/p>\n

                          [latex]\\displaystyle \\log _b \\left( {xy} \\right) = \\log _b \\left( x \\right) + \\log _b \\left( y \\right)[\/latex]<\/p>\n

                          If we apply this rule repeatedly we can devise another rule for simplifying expressions of the form [latex]\\log_b x^p[\/latex].<\/p>\n

                          Recall that [latex]x^p[\/latex]\u00a0can be thought of as [latex]x \\cdot x \\cdot x \\cdots x[\/latex]\u00a0where there are [latex]p[\/latex]\u00a0factors of [latex]x[\/latex]. Then we have:<\/p>\n

                          [latex]\\displaystyle \\begin{align} \\log_b(x^p) &= \\log_b (x \\cdot x \\cdots x) \\\\ &= \\log_b x + \\log_b x + \\cdots +\\log_b x \\\\ &= p\\log_b x \\end{align}[\/latex]<\/p>\n

                          Since the [latex]p[\/latex]\u00a0factors of [latex]x[\/latex]\u00a0are converted to [latex]p[\/latex]\u00a0summands by the product rule formula.<\/p>\n

                          Example 1: Simplify the expression\u00a0[latex]\\log_3(3^x\\cdot 9x^{100})[\/latex]<\/h3>\n

                          First expand the log:<\/p>\n

                          [latex]\\displaystyle \\log_3(3^x\\cdot 9x^{100}) =\\log_3 (3^x) + \\log_3 9 + \\log_3(x^{100})[\/latex]<\/p>\n

                          Next use the product and power rule to simplify:<\/p>\n

                          [latex]\\displaystyle \\log_3 (3^x) + \\log_3 9 + \\log_3 (x^{100})= x+2+100\\log_3 x[\/latex]<\/p>\n

                          Example 2: Solve [latex]2^{(x+1)}=10^3[\/latex]\u00a0for [latex]x[\/latex]\u00a0using logarithms<\/h3>\n

                          Start\u00a0by taking the logarithm with base [latex]2[\/latex]\u00a0of both sides:<\/p>\n

                          [latex]\\displaystyle \\begin{align} \\log_2 (2^{(x+1)}) &= \\log_2 (10^3)\\\\ x+1&=3\\log_2(10)\\\\ x&=3\\log_2(10)-1 \\end{align}[\/latex]<\/p>\n

                          Therefore a solution would be\u00a0[latex]x=3\\log_2(10) -1.[\/latex]<\/p>\n

                          Logarithms of Quotients<\/h2>\n

                          The logarithm of the ratio of two quantities is the difference of the logarithms of the quantities. In symbols, [latex]\\log_b\\left( \\frac{x}{y}\\right) = \\log_bx - \\log_by.[\/latex]<\/p>\n

                          \n

                          Learning Objectives<\/h3>\n

                          Relate the quotient rule for logarithms to the rules for operating with exponents, and use this rule to rewrite logarithms of quotients<\/p>\n<\/div>\n

                          \n

                          Key Takeaways<\/h3>\n

                          Key Points<\/h4>\n
                            \n
                          • The logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number.<\/li>\n
                          • The logarithm of a product is the sum of the logarithms of the factors.<\/li>\n
                          • The logarithm of the ratio or quotient of two numbers is the difference of the logarithms.<\/li>\n<\/ul>\n

                            Key Terms<\/h4>\n
                              \n
                            • exponent<\/strong>: The power to which a number, symbol, or expression is to be raised. For example, the 3 in [latex]x^3[\/latex].<\/li>\n<\/ul>\n<\/div>\n

                              We have already seen that the logarithm of a product is the sum of the logarithms of the factors:<\/p>\n

                              [latex]\\displaystyle \\log_b(xy) = \\log_bx + \\log_by[\/latex]<\/p>\n

                              Similarly, the logarithm of the ratio of two quantities is the difference of the logarithms:<\/p>\n

                              [latex]\\displaystyle \\log_b\\left( \\frac{x}{y}\\right) = \\log_bx - log_by.[\/latex]<\/p>\n

                              We can show that this is true by the following example:<\/p>\n

                              Let [latex]u=\\log_b x[\/latex]\u00a0and [latex]v=\\log_b y[\/latex].<\/p>\n

                              Then [latex]b^u=x[\/latex]\u00a0and [latex]b^v=y.[\/latex]<\/p>\n

                              Then:<\/p>\n

                              [latex]\\displaystyle \\begin{align} \\log_b\\left(\\frac{x}{y}\\right)&=\\log_b\\left({b^u \\over b^v}\\right)\\\\ &= \\log_b(b^{u-v}) \\\\ &=u-v\\\\ &= \\log_b x - \\log_b y \\end{align}[\/latex]<\/p>\n

                              Another way to show that this rule is true, is to apply both the power and product rules and the fact that dividing by [latex]y[\/latex]\u00a0is the same is multiplying by [latex]y^{-1}.[\/latex]\u00a0So we can write:<\/p>\n

                              [latex]\\displaystyle \\begin{align} \\log_b\\left(\\frac{x}{y}\\right)&=\\log_b(x\\cdot y^{-1})\\\\ & = \\log_bx + \\log_b(y^{-1})\\\\& = \\log_bx -\\log_by \\end{align}[\/latex]<\/p>\n

                              Example: write the expression [latex]\\log_2\\left({x^4y^9 \\over z^{100}}\\right)[\/latex]\u00a0in a simpler way<\/h3>\n

                              By applying the product, power, and quotient rules, you could write this expression as:<\/p>\n

                              [latex]\\displaystyle \\log_2(x^4)+\\log_2(y^9)-\\log_2(z^{100}) = 4\\log_2x+9\\log_2y-100\\log_2z.[\/latex]<\/p>\n

                              Changing Logarithmic Bases<\/h2>\n

                              A logarithm written in one base can be converted to an equal quantity written in a different base.<\/p>\n

                              \n

                              Learning Objectives<\/h3>\n

                              Use the change of base formula to convert logarithms to different bases<\/p>\n<\/div>\n

                              \n

                              Key Takeaways<\/h3>\n

                              Key Points<\/h4>\n
                                \n
                              • The base of a logarithm can be changed by expressing it as the quotient of two logarithms with a common\u00a0base.<\/li>\n
                              • Changing a logarithm’s base to [latex]10[\/latex] makes it much simpler to evaluate; it can be done on a calculator.<\/li>\n<\/ul>\n

                                Key Terms<\/h4>\n
                                  \n
                                • logarithm<\/strong>: The logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number.<\/li>\n
                                • base<\/strong>: A number raised to the power of an exponent.<\/li>\n<\/ul>\n<\/div>\n

                                  Most common scientific calculators have a key for computing logarithms with base [latex]10[\/latex], but do not have keys for other bases. So, if you needed to get an approximation to a number like [latex]\\log_4(9)[\/latex]\u00a0it can be difficult to do so. One could easily guess that it is between [latex]1[\/latex]\u00a0and [latex]2[\/latex]\u00a0since [latex]9[\/latex]\u00a0is between [latex]4^1[\/latex]\u00a0and [latex]4^2[\/latex], but it is difficult to get an accurate approximation. Fortunately, there is a change of base formula that can help.<\/p>\n

                                  Change of Base Formula<\/h3>\n

                                  The change of base formula for logarithms is:<\/p>\n

                                  [latex]\\displaystyle \\log_a(x)=\\frac{\\log_b(x)}{\\log_b(a)}[\/latex]<\/p>\n

                                  Thus, for example, we could calculate that [latex]\\log_4(9)=\\frac{\\log_{10}(9)}{\\log_{10}(4)}[\/latex]\u00a0which could be computed on almost any handheld calculator.<\/p>\n

                                  Deriving the Formula<\/h3>\n

                                  To see why the formula is true, give [latex]\\log_a(x)[\/latex]\u00a0a name like [latex]z[\/latex]:<\/p>\n

                                  [latex]\\displaystyle z=\\log_a(x)[\/latex]<\/p>\n

                                  Write this as\u00a0[latex]a^z=x[\/latex]<\/p>\n

                                  Now take the logarithm with base [latex]b[\/latex]\u00a0of both sides, yielding:<\/p>\n

                                  [latex]\\displaystyle \\log_b a^z = \\log_bx[\/latex]<\/p>\n

                                  Using the power rule gives:<\/p>\n

                                  [latex]\\displaystyle z \\cdot \\log_ba = \\log_b x[\/latex]<\/p>\n

                                  Dividing both sides by [latex]\\log_ba[\/latex]\u00a0gives:<\/p>\n

                                  [latex]\\displaystyle z={\\log_b x \\over \\log_ba}.[\/latex]<\/p>\n

                                  Thus we have\u00a0[latex]\\log_a x ={\\log_b x \\over \\log_b a}.[\/latex]<\/p>\n

                                  Example<\/h3>\n

                                  An expression of the form [latex]\\log_5(10^{x^2+1})[\/latex]\u00a0might be easier to graph on a graphing calculator or other device if it were written in base [latex]10[\/latex]\u00a0instead of base 5. The change-of-base formula can be applied to it:<\/p>\n

                                  [latex]\\displaystyle \\log_5(10^{x^2+1}) = {\\log_{10}(10^{x^2+1}) \\over \\log_{10}5}[\/latex]<\/p>\n

                                  Which can be written as\u00a0[latex]{x^2+1 \\over \\log_{10} 5}.[\/latex]<\/p>\n","protected":false},"author":473810,"menu_order":3,"template":"","meta":{"_candela_citation":"","CANDELA_OUTCOMES_GUID":"","pb_show_title":"","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-16178","chapter","type-chapter","status-publish","hentry"],"part":0,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapters\/16178","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/wp\/v2\/users\/473810"}],"version-history":[{"count":0,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapters\/16178\/revisions"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/parts\/0"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapters\/16178\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/wp\/v2\/media?parent=16178"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapter-type?post=16178"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/wp\/v2\/contributor?post=16178"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/wp\/v2\/license?post=16178"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}