Logarithms of Products

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, [latex]\log_b(xy)=\log_b(x)+\log_b(y).[/latex]

Logarithms

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]

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].

It is useful to think of logarithms as inverses of exponentials. So, for example:

[latex]\displaystyle \log_b(b^z)=z[/latex]

And:

[latex]\displaystyle b^{\log_b(z)}=z[/latex]

Product Rule for Logarithms

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:

[latex]\displaystyle log_b(xy) = log_b(x) + log_b(y)[/latex]

We can see that this rule is true by writing the logarithms in terms of exponentials.

Let [latex]\log_b(x)=v[/latex] and [latex]\log_b(y)=w.[/latex]

Writing these equations as exponentials:

[latex]\displaystyle b^v=x[/latex]

And:

[latex]\displaystyle b^w=y.[/latex]

Then note that:

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

Taking the logarithm base [latex]b[/latex] of both sides of this last equation yields:

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

This is a very useful property of logarithms, because it can sometimes simplify more complex expressions. For example:

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

Then because [latex]100[/latex] is [latex]10^2[/latex], we have:

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

Logarithms of Powers

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, [latex]\log_b(x^p)=p\log_b(x).[/latex]

The Power Rule for Logarithms

We have already seen that the logarithm of a product is the sum of the logarithms of the factors:

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

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

Recall that [latex]x^p[/latex] can be thought of as [latex]x \cdot x \cdot x \cdots x[/latex] where there are [latex]p[/latex] factors of [latex]x[/latex]. Then we have:

[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]

Since the [latex]p[/latex] factors of [latex]x[/latex] are converted to [latex]p [/latex] summands by the product rule formula.

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

First expand the log:

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

Next use the product and power rule to simplify:

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

Example 2: Solve [latex]2^{(x+1)}=10^3[/latex] for [latex]x[/latex] using logarithms

Start by taking the logarithm with base [latex]2[/latex] of both sides:

[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]

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

Logarithms of Quotients

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]

We have already seen that the logarithm of a product is the sum of the logarithms of the factors:

[latex]\displaystyle \log_b(xy) = \log_bx + \log_by[/latex]

Similarly, the logarithm of the ratio of two quantities is the difference of the logarithms:

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

We can show that this is true by the following example:

Let [latex]u=\log_b x[/latex] and [latex]v=\log_b y[/latex].

Then [latex]b^u=x[/latex] and [latex]b^v=y.[/latex]

Then:

[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]

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] is the same is multiplying by [latex]y^{-1}.[/latex] So we can write:

[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]

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

By applying the product, power, and quotient rules, you could write this expression as:

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

Changing Logarithmic Bases

A logarithm written in one base can be converted to an equal quantity written in a different base.

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] it can be difficult to do so. One could easily guess that it is between [latex]1[/latex] and [latex]2[/latex] since [latex]9[/latex] is between [latex]4^1[/latex] and [latex]4^2[/latex], but it is difficult to get an accurate approximation. Fortunately, there is a change of base formula that can help.

Change of Base Formula

The change of base formula for logarithms is:

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

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

Deriving the Formula

To see why the formula is true, give [latex]\log_a(x)[/latex] a name like [latex]z[/latex]:

[latex]\displaystyle z=\log_a(x)[/latex]

Write this as [latex]a^z=x[/latex]

Now take the logarithm with base [latex]b[/latex] of both sides, yielding:

[latex]\displaystyle \log_b a^z = \log_bx[/latex]

Using the power rule gives:

[latex]\displaystyle z \cdot \log_ba = \log_b x[/latex]

Dividing both sides by [latex]\log_ba[/latex] gives:

[latex]\displaystyle z={\log_b x \over \log_ba}.[/latex]

Thus we have [latex]\log_a x ={\log_b x \over \log_b a}. [/latex]

Example

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

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

Which can be written as [latex]{x^2+1 \over \log_{10} 5}. [/latex]