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Section 14.1 Combining Logarithms

You might remember that we have exponent rules that tell us how we can combine things with exponents. Now, we know what logarithms are, which are just the inverse of exponentials. So, we will have similar rules for logs, but with the insides/outsides flipped.
Table 14.1. Log and Exponent Properties
Exponent Rule Logarithm Rule
\(b^xb^y=b^{x+y}\) \(\log_b(x)+\log_b(y)=\log_b(xy)\)
Multiply outside, add inside Add outside, multiply inside
\(\dfrac{b^x}{b^y}=b^{x-y}\) \(\log_b(x)-\log_b(y)=\log_b\left(\dfrac{x}{y}\right)\)
Divide outside, subtract inside Subtract outside, divide inside
\(\left(b^x\right)^y=b^{xy}\) \(y\log_b(x)=\log_b(x^y)\)
Exponent outside, multiply inside Multiply outside, exponent inside

Example 14.2.

Suppose we want to combine the following expression into one logarithm:

\begin{equation*} 5\log_2(x)-3\log_2(y)+\frac{1}{2}\log_2(x+1) \end{equation*}

We will need to use the first two rules in the table to combine the logs, but those rules don't work if there are numbers out front. So, we need to use the third rule in the table to deal with those first. We can bring the numbers inside the logs, but they have to go in the exponent.

\begin{equation*} \log_2\left(x^5\right)-\log_2\left(y^3\right)+\log_2\left((x+1)^{\frac{1}{2}}\right) \end{equation*}

Notice that in the last one we had to put parentheses around the \(x+1\text{.}\) The reason is that when we bring the exponent inside, it has to be an exponent on the entire inside. Without the parentheses, the \(\frac{1}{2}\) power would only be on the 1, completely missing the \(x\text{.}\)

Now we are ready to combine the logs into one. Since the log with \(y^3\) is being subtracted, our second rule tells us that the \(y^3\) will go in the denominator of our fraction.

\begin{equation*} \log_2\left(\dfrac{x^5(x+1)^{\frac{1}{2}}}{y^3}\right) \end{equation*}

Notice that when we brought the exponents inside, the log with the \(y\) only brought the 3 inside, but left the negative. If you prefer, you can bring the negative inside, too. You'll just have some negative exponents to deal with, but because those move things to the denominator, you'll end up with the same answer.

\begin{gather*} 5\log_2(x)-3\log_2(y)+\frac{1}{2}\log_2(x+1)\\ \log_2\left(x^5\right)+\log_2\left(y^{-3}\right)+\log_2\left((x+1)^{\frac{1}{2}}\right)\\ \log_2\left(x^5y^{-3}(x+1)^{\frac{1}{2}}\right)\\ \log_2\left(\dfrac{x^5(x+1)^{\frac{1}{2}}}{y^3}\right) \end{gather*}

Checkpoint 14.3.

Combine the following into one logarithm

\begin{equation*} 3\ln(x-2)-\frac{3}{2}\ln(y)-7\ln(z) \end{equation*}
Answer.
\begin{equation*} \log_2\left(\dfrac{(x-2)^3}{y^{\frac{3}{2}}z^7}\right) \end{equation*}
Solution.

We will need to use the first two rules in the table to combine the logs, but those rules don't work if there are numbers out front. So, we need to use the third rule in the table to deal with those first. We can bring the numbers inside the logs, but they have to go in the exponent.

\begin{equation*} \ln\left((x-2)^3\right)-\ln\left(y^{\frac{3}{2}}\right)-\ln\left(z^7\right) \end{equation*}

Now we are ready to combine the logs into one. Since the log with \(y^{\frac{3}{2}}\) and the \(z^7\) are being subtracted, our second rule tells us that those will go in the denominator of our fraction.

\begin{equation*} \log_2\left(\dfrac{(x-2)^3}{y^{\frac{3}{2}}z^7}\right) \end{equation*}