Expanding and Condensing Logarithms

HOW TO

Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm.

1. Apply the power property first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power.
2. Next apply the product property. Rewrite sums of logarithms as the logarithm of a product.
3. Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient.

Example 9

Using the Product and Quotient Rules to Combine Logarithms

Write $log_3(5)+log_3(8)−log_3(2)$ as a single logarithm.

Solution

Using the product and quotient rules

$log_3(5)+log_3(8)=log_3(5⋅8)=log_3(40)$

This reduces our original expression to

$log_3(40)−log_3(2)$

Then, using the quotient rule

$log_3(40)−log_3(2)=log_3 \left(\frac{40}{2}\right)=log_3(20)$

Try It #9

Condense $log3−log4+log5−log6$.

Example 10

Condensing Complex Logarithmic Expressions

Condense $log_2(x^2)+\frac{1}{2}log_2(x−1)−3log_2((x+3)^2)$.

Solution

We apply the power rule first:

$\log _{2}\left(x^{2}\right)+\frac{1}{2} \log _{2}(x-1)-3 \log _{2}\left((x+3)^{2}\right)=\log _{2}\left(x^{2}\right)+\log _{2}(\sqrt{x-1})-\log _{2}\left((x+3)^{6}\right)$

Next we apply the product rule to the sum:

$\log _{2}\left(x^{2}\right)+\log _{2}(\sqrt{x-1})-\log _{2}\left((x+3)^{6}\right)=\log _{2}\left(x^{2} \sqrt{x-1}\right)-\log _{2}\left((x+3)^{6}\right)$

Finally, we apply the quotient rule to the difference:

$\log _{2}\left(x^{2} \sqrt{x-1}\right)-\log _{2}\left((x+3)^{6}\right)=\log _{2} \frac{x^{2} \sqrt{x-1}}{(x+3)^{6}}$

Try It #10

Rewrite $log(5)+0.5log(x)−log(7x−1)+3log(x−1)$ as a single logarithm.

Example 11

Rewriting as a Single Logarithm

Rewrite $2logx−4log(x+5)+ \frac{1}{x}log(3x+5)$ as a single logarithm.

Solution

We apply the power rule first:

$2 \log x-4 \log (x+5)+\frac{1}{x} \log (3 x+5)=\log \left(x^{2}\right)-\log (x+5)^{4}+\log \left((3 x+5)^{x^{-1}}\right)$

Next we rearrange and apply the product rule to the sum:

$\begin{gathered} \log \left(x^{2}\right)-\log (x+5)^{4}+\log \left((3 x+5)^{x^{-1}}\right) \\ =\log \left(x^{2}\right)+\log \left((3 x+5)^{x^{-1}}\right)-\log (x+5)^{4} \\ =\log \left(x^{2}(3 x+5)^{x^{-1}}\right)-\log (x+5)^{4} \end{gathered}$

Finally, we apply the quotient rule to the difference:

$=\log \left(x^{2}(3 x+5)^{x^{-1}}\right)-\log (x+5)^{4}=\log \frac{x^{2}(3 x+5)^{x^{-1}}}{(x+5)^{4}}$

Try It #11

Condense $4(3 \log (x)+\log (x+5)-\log (2 x+3))$

Example 12

Applying of the Laws of Logs

Recall that, in chemistry, $\mathrm{pH}=-\log \left[H^{+}\right]$. If the concentration of hydrogen ions in a liquid is doubled, what is the effect on $\mathrm{pH}$?

Solution

Suppose $C$ is the original concentration of hydrogen ions, and $P$ is the original $\mathrm{pH}$ of the liquid. Then $P=-\log (C)$. If the concentration is doubled, the new concentration is $2 C$. Then the $\mathrm{pH}$ of the new liquid is

$\mathrm{pH}=-\log (2 C)$

Using the product rule of logs

$\mathrm{pH}=-\log (2 C)=-(\log (2)+\log (C))=-\log (2)-\log (C)$

Since $P=-\log (C)$, the new $\mathrm{pH}$ is

$\mathrm{pH}=P-\log (2) \approx P-0.301$

When the concentration of hydrogen ions is doubled, the $\mathrm{pH}$ decreases by about $0.301$.

Try It #12

How does the $\mathrm{pH}$ change when the concentration of positive hydrogen ions is decreased by half?