{"id":11230,"date":"2015-07-14T19:07:25","date_gmt":"2015-07-14T19:07:25","guid":{"rendered":"https:\/\/courses.candelalearning.com\/osprecalc\/?post_type=chapter&p=11230"},"modified":"2015-09-10T19:00:06","modified_gmt":"2015-09-10T19:00:06","slug":"use-the-definition-of-a-logarithm-to-solve-logarithmic-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/use-the-definition-of-a-logarithm-to-solve-logarithmic-equations\/","title":{"raw":"Use the definition of a logarithm to solve logarithmic equations","rendered":"Use the definition of a logarithm to solve logarithmic equations"},"content":{"raw":"

We have already seen that every logarithmic equation<\/strong> [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex] is equivalent to the exponential equation [latex]{b}^{y}=x[\/latex]. We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.<\/p>\r\n

For example, consider the equation [latex]{\\mathrm{log}}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x - 5\\right)=3[\/latex]. To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for x<\/em>:<\/p>\r\n\r\n

[latex]\\begin{cases}{\\mathrm{log}}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x - 5\\right)=3\\hfill & \\hfill \\\\ \\text{ }{\\mathrm{log}}_{2}\\left(2\\left(3x - 5\\right)\\right)=3\\hfill & \\text{Apply the product rule of logarithms}.\\hfill \\\\ \\text{ }{\\mathrm{log}}_{2}\\left(6x - 10\\right)=3\\hfill & \\text{Distribute}.\\hfill \\\\ \\text{ }{2}^{3}=6x - 10\\hfill & \\text{Apply the definition of a logarithm}.\\hfill \\\\ \\text{ }8=6x - 10\\hfill & \\text{Calculate }{2}^{3}.\\hfill \\\\ \\text{ }18=6x\\hfill & \\text{Add 10 to both sides}.\\hfill \\\\ \\text{ }x=3\\hfill & \\text{Divide by 6}.\\hfill \\end{cases}[\/latex]<\/div>\r\n
\r\n

A General Note: Using the Definition of a Logarithm to Solve Logarithmic Equations<\/h3>\r\n

For any algebraic expression S<\/em> and real numbers b<\/em> and c<\/em>, where [latex]b>0,\\text{ }b\\ne 1[\/latex],<\/p>\r\n\r\n

[latex]{\\mathrm{log}}_{b}\\left(S\\right)=c\\text{if and only if}{b}^{c}=S[\/latex]<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n

Example 9: Using Algebra to Solve a Logarithmic Equation<\/h3>\r\n

Solve [latex]2\\mathrm{ln}x+3=7[\/latex].<\/p>\r\n\r\n<\/div>\r\n

\r\n
\r\n

Solution<\/h3>\r\n

[latex]\\begin{cases}2\\mathrm{ln}x+3=7\\hfill & \\hfill \\\\ \\text{ }2\\mathrm{ln}x=4\\hfill & \\text{Subtract 3}.\\hfill \\\\ \\text{ }\\mathrm{ln}x=2\\hfill & \\text{Divide by 2}.\\hfill \\\\ \\text{ }x={e}^{2}\\hfill & \\text{Rewrite in exponential form}.\\hfill \\end{cases}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n

Try It 9<\/h3>\r\n

Solve [latex]6+\\mathrm{ln}x=10[\/latex].<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>\r\n

\r\n
\r\n
\r\n

Example 10: Using Algebra Before and After Using the Definition of the Natural Logarithm<\/h3>\r\n

Solve [latex]2\\mathrm{ln}\\left(6x\\right)=7[\/latex].<\/p>\r\n\r\n<\/div>\r\n

\r\n
\r\n

Solution<\/h3>\r\n

[latex]\\begin{cases}2\\mathrm{ln}\\left(6x\\right)=7\\hfill & \\hfill \\\\ \\text{ }\\mathrm{ln}\\left(6x\\right)=\\frac{7}{2}\\hfill & \\text{Divide by 2}.\\hfill \\\\ \\text{ }6x={e}^{\\left(\\frac{7}{2}\\right)}\\hfill & \\text{Use the definition of }\\mathrm{ln}.\\hfill \\\\ \\text{ }x=\\frac{1}{6}{e}^{\\left(\\frac{7}{2}\\right)}\\hfill & \\text{Divide by 6}.\\hfill \\end{cases}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n

Try It 10<\/h3>\r\n

Solve [latex]2\\mathrm{ln}\\left(x+1\\right)=10[\/latex].<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>\r\n

\r\n
\r\n
\r\n

Example 11: Using a Graph to Understand the Solution to a Logarithmic Equation<\/h3>\r\n

Solve [latex]\\mathrm{ln}x=3[\/latex].<\/p>\r\n\r\n<\/div>\r\n

\r\n
\r\n

Solution<\/h3>\r\n

[latex]\\begin{cases}\\mathrm{ln}x=3\\hfill & \\hfill \\\\ x={e}^{3}\\hfill & \\text{Use the definition of the natural logarithm}\\text{.}\\hfill \\end{cases}[\/latex]<\/p>\r\n

Figure 2\u00a0represents the graph of the equation. On the graph, the x<\/em>-coordinate of the point at which the two graphs intersect is close to 20. In other words [latex]{e}^{3}\\approx 20[\/latex]. A calculator gives a better approximation: [latex]{e}^{3}\\approx 20.0855[\/latex].<\/p>\r\n\r\n

\"Graph<\/figure>
\r\n
Figure 2.<\/strong> The graphs of [latex]y=\\mathrm{ln}x[\/latex] and y\u00a0<\/em>= 3 cross at the point [latex]\\left(e^3,3\\right)[\/latex], which is approximately (20.0855, 3).<\/div>\r\n<\/figcaption><\/figure><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n

Try It 11<\/h3>\r\n

Use a graphing calculator to estimate the approximate solution to the logarithmic equation [latex]{2}^{x}=1000[\/latex] to 2 decimal places.<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>","rendered":"

We have already seen that every logarithmic equation<\/strong> [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex] is equivalent to the exponential equation [latex]{b}^{y}=x[\/latex]. We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.<\/p>\n

For example, consider the equation [latex]{\\mathrm{log}}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x - 5\\right)=3[\/latex]. To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for x<\/em>:<\/p>\n

[latex]\\begin{cases}{\\mathrm{log}}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x - 5\\right)=3\\hfill & \\hfill \\\\ \\text{ }{\\mathrm{log}}_{2}\\left(2\\left(3x - 5\\right)\\right)=3\\hfill & \\text{Apply the product rule of logarithms}.\\hfill \\\\ \\text{ }{\\mathrm{log}}_{2}\\left(6x - 10\\right)=3\\hfill & \\text{Distribute}.\\hfill \\\\ \\text{ }{2}^{3}=6x - 10\\hfill & \\text{Apply the definition of a logarithm}.\\hfill \\\\ \\text{ }8=6x - 10\\hfill & \\text{Calculate }{2}^{3}.\\hfill \\\\ \\text{ }18=6x\\hfill & \\text{Add 10 to both sides}.\\hfill \\\\ \\text{ }x=3\\hfill & \\text{Divide by 6}.\\hfill \\end{cases}[\/latex]<\/div>\n
\n

A General Note: Using the Definition of a Logarithm to Solve Logarithmic Equations<\/h3>\n

For any algebraic expression S<\/em> and real numbers b<\/em> and c<\/em>, where [latex]b>0,\\text{ }b\\ne 1[\/latex],<\/p>\n

[latex]{\\mathrm{log}}_{b}\\left(S\\right)=c\\text{if and only if}{b}^{c}=S[\/latex]<\/div>\n<\/div>\n
\n
\n
\n

Example 9: Using Algebra to Solve a Logarithmic Equation<\/h3>\n

Solve [latex]2\\mathrm{ln}x+3=7[\/latex].<\/p>\n<\/div>\n

\n
\n

Solution<\/h3>\n

[latex]\\begin{cases}2\\mathrm{ln}x+3=7\\hfill & \\hfill \\\\ \\text{ }2\\mathrm{ln}x=4\\hfill & \\text{Subtract 3}.\\hfill \\\\ \\text{ }\\mathrm{ln}x=2\\hfill & \\text{Divide by 2}.\\hfill \\\\ \\text{ }x={e}^{2}\\hfill & \\text{Rewrite in exponential form}.\\hfill \\end{cases}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

\n

Try It 9<\/h3>\n

Solve [latex]6+\\mathrm{ln}x=10[\/latex].<\/p>\n

Solution<\/a><\/p>\n<\/div>\n

\n
\n
\n

Example 10: Using Algebra Before and After Using the Definition of the Natural Logarithm<\/h3>\n

Solve [latex]2\\mathrm{ln}\\left(6x\\right)=7[\/latex].<\/p>\n<\/div>\n

\n
\n

Solution<\/h3>\n

[latex]\\begin{cases}2\\mathrm{ln}\\left(6x\\right)=7\\hfill & \\hfill \\\\ \\text{ }\\mathrm{ln}\\left(6x\\right)=\\frac{7}{2}\\hfill & \\text{Divide by 2}.\\hfill \\\\ \\text{ }6x={e}^{\\left(\\frac{7}{2}\\right)}\\hfill & \\text{Use the definition of }\\mathrm{ln}.\\hfill \\\\ \\text{ }x=\\frac{1}{6}{e}^{\\left(\\frac{7}{2}\\right)}\\hfill & \\text{Divide by 6}.\\hfill \\end{cases}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

\n

Try It 10<\/h3>\n

Solve [latex]2\\mathrm{ln}\\left(x+1\\right)=10[\/latex].<\/p>\n

Solution<\/a><\/p>\n<\/div>\n

\n
\n
\n

Example 11: Using a Graph to Understand the Solution to a Logarithmic Equation<\/h3>\n

Solve [latex]\\mathrm{ln}x=3[\/latex].<\/p>\n<\/div>\n

\n
\n

Solution<\/h3>\n

[latex]\\begin{cases}\\mathrm{ln}x=3\\hfill & \\hfill \\\\ x={e}^{3}\\hfill & \\text{Use the definition of the natural logarithm}\\text{.}\\hfill \\end{cases}[\/latex]<\/p>\n

Figure 2\u00a0represents the graph of the equation. On the graph, the x<\/em>-coordinate of the point at which the two graphs intersect is close to 20. In other words [latex]{e}^{3}\\approx 20[\/latex]. A calculator gives a better approximation: [latex]{e}^{3}\\approx 20.0855[\/latex].<\/p>\n

\"Graph<\/figure>\n
\nFigure 2.<\/strong> The graphs of [latex]y=\\mathrm{ln}x[\/latex] and y\u00a0<\/em>= 3 cross at the point [latex]\\left(e^3,3\\right)[\/latex], which is approximately (20.0855, 3).<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
\n

Try It 11<\/h3>\n

Use a graphing calculator to estimate the approximate solution to the logarithmic equation [latex]{2}^{x}=1000[\/latex] to 2 decimal places.<\/p>\n

Solution<\/a><\/p>\n<\/div>\n\n\t\t\t

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