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Section 17.2 Using Log and Exponenetial Rules

Sometimes, you might have an equation that have multiple logs in them. When that happens, we could either use exponential rules or logarithm rules to help us solve the equation.

Example 17.6.

Suppose we want to solve the following equation:

\begin{equation*} \log_6(x+5)+\log_6(3)=2 \end{equation*}

Depending on whether you prefer working with exponent rules or log rules, there are two ways you could approach this problem. We will show you both, and you can pick the one that makes the most sense to you.

Option 1: Log Rules. In Section 16.1, we learned about how to take two logs that are added and combine them into one. So, we will start by comining the \(\log_6(x+5)\) and the \(\log_6(3)\text{.}\)

\begin{equation*} \log_6\left(3(x+5)\right)=2 \end{equation*}

Now that we only have one log, and everything is inside of it, we can use an exponential base 6 to get rid of it.

\begin{align*} \log_6\left(3(x+5)\right)\amp=2\\ {\color{red}{6}}^{\log_6\left(3(x+5)\right)}\amp={\color{red}{6}}^2\\ 3(x+5)\amp=36 \end{align*}

Now that we've gotten rid of the log, we can solve the rest of the equation using usual order of operations.

\begin{align*} 3(x+5)\amp=36\\ \frac{3(x+5)}{\color{red}{3}}\amp=\frac{36}{\color{red}{3}}\\ x+5\amp=12\\ x+5{\color{red}{-5}}\amp=12{\color{red}{-5}}\\ x\amp=7 \end{align*}

Option 2: Exponent Rules. Suppose we forgot our log rules, and we want to jump right into getting \(x\) by itself. In that case, we'll need to use our exponent rules eventually to deal with the two logs. In the meantime, we notice that \(\log_6(3)\) is just some number, so we can just subtract to move it to the other side.

\begin{align*} \log_6(x+5)+\log_6(3)\amp=2\\ \log_6(x+5)+\log_6(3){\color{red}{-\log_6(3)}}\amp=2{\color{red}{-\log_6(3)}}\\ \log_6(x+5)\amp= 2-\log_6(3) \end{align*}

We are now ready to use an exponential base 6 to cancel the log base 6 that's on the left side. Although we do want it to eventually cancel with the log base 6 on the right side, we will need to use exponent rules to get the 2 out of the way first.

\begin{align*} {\color{red}{6}}^{\log_6(x+5)}\amp= {\color{red}{6}}^{2-\log_6(3)}\\ x+5\amp=\dfrac{6^2}{6^{\log_6(3)}}\\ x+5\amp=\frac{36}{3}\\ x+5\amp=12 \end{align*}

Finally, we can subtract 5 from both sides to get our answer of \(x=7\)/

Checkpoint 17.7.

Solve the fllowing equation:

\begin{equation*} \log_5(x+1)-\log_5(3)=2 \end{equation*}
Answer.

\(x=76\)

Solution.

Depending on whether you prefer working with exponent rules or log rules, there are two ways you could approach this problem. We will show you both, and you can pick the one that makes the most sense to you.

Option 1: Log Rules. In Section 16.1, we learned about how to take two logs that are added and combine them into one. So, we will start by comining the \(\log_5(x-1)\) and the \(\log_5(3)\text{.}\)

\begin{equation*} \log_5\left(\frac{x-1}{3}\right)=2 \end{equation*}

Now that we only have one log, and everything is inside of it, we can use an exponential base 5 to get rid of it.

\begin{align*} \log_5\left(\frac{x-1}{3}\right)\amp=2\\ {\color{red}{5}}^{\log_5\left({x-1}{3}\right)}\amp={\color{red}{5}}^2\\ \frac{x-1}{3}\amp=25 \end{align*}

Now that we've gotten rid of the log, we can solve the rest of the equation using usual order of operations.

\begin{align*} \frac{x-1}{3}\amp=25\\ {\color{red}{3\cdot}}\frac{x-1}{3}\amp={\color{red}{3\cdot}}25\\ x-1\amp=75\\ x-1{\color{red}{+1}}\amp=75{\color{red}{+1}}\\ x\amp=76 \end{align*}

Option 2: Exponent Rules. Suppose we forgot our log rules, and we want to jump right into getting \(x\) by itself. In that case, we'll need to use our exponent rules eventually to deal with the two logs. In the meantime, we notice that \(\log_5(3)\) is just some number, so we can just add to move it to the other side.

\begin{align*} \log_5(x-1)-\log_5(3)\amp=2\\ \log_5(x-1)-\log_5(3){\color{red}{+\log_5(3)}}\amp=2{\color{red}{+\log_5(3)}}\\ \log_5(x-1)\amp= 2+\log_5(3) \end{align*}

We are now ready to use an exponential base 6 to cancel the log base 5 that's on the left side. Although we do want it to eventually cancel with the log base 5 on the right side, we will need to use exponent rules to get the 2 out of the way first.

\begin{align*} {\color{red}{5}}^{\log_5(x-1)}\amp= {\color{red}{5}}^{2+\log_5(3)}\\ x-1\amp=5^2 \cdot 5^{\log_5(3)}\\ x-1\amp=25\cdot 3\\ x-1\amp=75 \end{align*}

Finally, we can add 1 to both sides to get our answer of \(x=76\)/