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Section 17.3 (*)Solving Complicated Equations

This section is still under construction

The most complicated exponential equations that we will solve in this class are ones with the variable on both sides of the equation. In that case, we need to use a log to get the variables out of the exponent. After that, we can combine the x's and solve. Technically, you can use any log that you want to solve these equations. Some will just be more useful than others, in some cases.

Example 17.8.

Suppose we want to solve the following equation:

\begin{equation*} 9^{4x+7}=81^{3x-1} \end{equation*}

There are two ways we might solve this, and you may choose any that you want.

Option 1: Exponent Ruels. Notice that \(81=9^2\text{.}\) So, we can use that to rewrite our equation:

\begin{equation*} 9^{4x+7}=\left(9^2\right)^{3x-1}=9^{2(3x-1)}=9^{6x-2} \end{equation*}

Now that both sides are exponentials base 9, we can see that using a log base 9 will be the most useful, since it will cancel both sides.

\begin{align*} {\color{red}{\log_9}}\left(9^{4x+7}\right)\amp={\color{red}{\log_9}}\left(9^{6x-2}\right)\\ 4x+7\amp=6x-2 \end{align*}

Now, we have no more logs or exponentials, so we go back to the solving strategies that we learned, for example, in Section 6.1.

\begin{align*} 4x+7\amp=6x-2\\ 4x+7{\color{red}{-4x}}\amp=6x-2{\color{red}{-4x}}\\ 7\amp=2x-2\\ 7{\color{red}{+2}}\amp=2x-2{\color{red}{+2}}\\ 9\amp=2x\\ \frac{9}{\color{red}{2}}\amp=\frac{2x}{\color{red}{2}}\\ \frac{9}{2}\amp= x \end{align*}

Option 2: Log Rules. We still notice that \(81=9^2\text{,}\) but instead of subsituting, that information just helps us decide that we want to take log base 9 for this problem. Although it will cancel on the left-hand side, it won't yet cancel on the right. Luckily, we can use our log properties to bring down the exponent and then simplify.

\begin{align*} {\color{red}{\log_9}}\left(9^{4x+7}\right)\amp={\color{red}{\log_9}}\left(81^{3x-1}\right)\\ 4x+7\amp=\log_9\left(81^{3x-1}\right)\\ 4x+7\amp=(3x-1)\log_9(81)\\ 4x+7\amp=(3x-1)\log_9\left(9^2\right)\\ 4x+7\amp=(3x-1)2\\ 4x+7\amp=6x-2 \end{align*}

Now we have gotten rid of the exponents and logs, so we can combine the x's and finish solving this equation.

\begin{align*} 4x+7{\color{red}{-4x}}\amp=6x-2{\color{red}{-4x}}\\ 4x+7{\color{red}{-4x}}\amp=6x-2{\color{red}{-4x}}\\ 7\amp=2x-2\\ 7{\color{red}{+2}}\amp=2x-2{\color{red}{+2}}\\ 9\amp=2x\\ \frac{9}{\color{red}{2}}\amp=\frac{2x}{\color{red}{2}}\\ \frac{9}{2}\amp= x \end{align*}

Checkpoint 17.9.

Solve the following equation for \(x\text{.}\)

\begin{equation*} 5^{x-3}=125^{x+7} \end{equation*}
Answer.

\(x=-12\)

Solution.

There are two ways we might solve this, and you may choose any that you want.

Option 1: Exponent Ruels. Notice that \(125=5^3\text{.}\) So, we can use that to rewrite our equation:

\begin{equation*} 5^{x-3}=\left(5^3\right)^{x+7}=5^{3(x+7)}=5^{3x+21} \end{equation*}

Now that both sides are exponentials base 5, we can see that using a log base 5 will be the most useful, since it will cancel both sides.

\begin{align*} {\color{red}{\log_5}}\left(5^{x-3}\right)\amp={\color{red}{\log_5}}\left(5^{3x+21}\right)\\ x-3\amp=3x+21 \end{align*}

Now, we have no more logs or exponentials, so we go back to the solving strategies that we learned, for example, in Section 6.1.

\begin{align*} x-3\amp=3x+21\\ x-3{\color{red}{-x}}\amp=3x+21{\color{red}{-x}}\\ -3\amp=2x+21\\ -3{\color{red}{-21}}\amp=2x+21{\color{red}{-21}}\\ -24\amp=2x\\ \frac{-24}{\color{red}{2}}\amp=\frac{2x}{\color{red}{2}}\\ -12\amp= x \end{align*}

Option 2: Log Rules. We still notice that \(125=5^3\text{,}\) but instead of subsituting, that information just helps us decide that we want to take log base 5 for this problem. Although it will cancel on the left-hand side, it won't yet cancel on the right. Luckily, we can use our log properties to bring down the exponent and then simplify.

\begin{align*} {\color{red}{\log_5}}\left(5^{x-3}\right)\amp={\color{red}{\log_5}}\left(125^{x+7}\right)\\ x-3\amp=\log_5\left(125^{x+7}\right)\\ x-3\amp=(x+7)\log_5(125)\\ x-3\amp=(x+7)\log_5\left(5^3\right)\\ x-3\amp=(x+7)3\\ x-3\amp=3x+21 \end{align*}

Now we have gotten rid of the exponents and logs, so we can combine the x's and finish solving this equation.

\begin{align*} x-3\amp=3x+21\\ x-3{\color{red}{-x}}\amp=3x+21{\color{red}{-x}}\\ -3\amp=2x+21\\ -3{\color{red}{-21}}\amp=2x+21{\color{red}{-21}}\\ -24\amp=2x\\ \frac{-24}{\color{red}{2}}\amp=\frac{2x}{\color{red}{2}}\\ -12\amp= x \end{align*}

Unfortunately, sometimes we don't have nice bases that will all cancel. In that case, there will be some logs in our answer. So, it will be heplful to keep track of which things are numbers, even though they don't look like it.

Example 17.10.

Suppose we want to solve the following equation:

\begin{equation*} 5^{x+3}=7^{2x-1} \end{equation*}

Unfortunately, since 5 is not a power of 7 and 7 is not a power of 5, there is no log we can pick that will cancel everything out. So, we have three options:

  1. Use log base 5: This would cancel the 5, but not the 7, so we would have a \(\log_5(7)\) floating around in our answer. If we are writing it on paper (like on a quiz or exam), that's no problem, but we can't enter bases for our logs into WeBWorK.

  2. Use log base 7: This would cancel the 7, but not the 5, so we would have a \(\log_7(5)\) floating around in our answer. If we are writing it on paper (like on a quiz or exam), that's no problem, but we can't enter bases for our logs into WeBWorK.

  3. Use natural log: This won't cancel the 5 or the 7, so we will have a \(\ln(5)\) and a \(\ln(7)\) in our answer. While that may be more to keep track of while solving the equation, the benefit is that we can type natural log into WeBWorK directly. The other benefit is that you don't have to think about it: you can just always use natural log, no matter what the bases are.

As you can see, there are benefits and drawbacks to each option. Any of them are totally fine, and even professional mathematicians have different preferences! Pick whichever makes the most sense to you. In this example, we will show both using log base 5 and natural log. The solution with log base 7 is very similar to the solution with log base 5, so we leave that to you to work out.

Option 1: Log base 5. We will start by taking the log base 5 of both sides, canceling what we can, and then using log rules to bring down the remaining exponents.

\begin{align*} {\color{red}{\log_5}}\left(5^{x+3}\right)\amp={\color{red}{\log_5}}\left(7^{2x-1}\right)\\ x+3\amp=(2x-1)\log_5(7) \end{align*}

This is where it becomes really important to look carefully at what we have. On one side we have a nice little \(x+3\text{.}\) On the other, we have this \(\log_5(7)\) floating around. Although it doesn't look like it right away, that's just a number. I could plug that into a calculator and it would spit out some kind of decimal. So, we will treat it like any other number.

\begin{align*} x+3\amp=(2x-1)\log_5(7)\\ x+3\amp=2\log_5(7)x-\log_5(7)\\ x+3{\color{red}{-x}}\amp=2\log_5(7)x-\log_5(7){\color{red}{-x}}\\ 3\amp=2\log_5(7)x-x-\log_5(7)\\ 3{\color{red}{+\log_5(7)}}\amp=2\log_5(7)x-x-\log_5(7){\color{red}{+\log_5(7)}}\\ 3+\log_5(7)\amp=2\log_5(7)x-x\\ 3+\log_5(7)\amp=(2\log_5(7)-1)x\\ \frac{3+\log_5(7)}{\color{red}{2\log_5(7)-1}}\amp=\frac{(2\log_5(7)-1)x}{\color{red}{2\log_5(7)-1}}\\ \frac{3+\log_5(7)}{2\log_5(7)-1}\amp=x \end{align*}

Option 3: Natural log. We will start by taking the natural log of both sides and using log rules to bring the exponents down.

\begin{align*} 5^{x+3}\amp=7^{2x-1}\\ {\color{red}{\ln}}\left(5^{x+3}\right)\amp={\color{red}{\ln}}\left(7^{2x-1}\right)\\ (x+3)\ln(5)\amp=(2x-1)\ln(7) \end{align*}

Again, we have to pay close attention here to what is a number. Although \(\ln(5)\) and \(\ln(7)\) don't immediately look like numbers, there are no variables in them. I can plug each of them into a calculator and it will spit out some ugly-looking decimal. So, I will treat each of them the same way I would any other number.

\begin{align*} (x+3)\ln(5)\amp=(2x-1)\ln(7)\\ \ln(5)x+3\ln(5)\amp=2\ln(7)x-\ln(7)\\ \ln(5)x+3\ln(5){\color{red}{-2\ln(7)x}}\amp=2\ln(7)x-\ln(7){\color{red}{-2\ln(7)x}}\\ \ln(5)x+3\ln(5)-2\ln(7)x \amp= -\ln(7)\\ \ln(5)x+3\ln(5)-2\ln(7)x {\color{red}{-3\ln(5)}}\amp= -\ln(7){\color{red}{-3\ln(5)}}\\ \ln(5)x-2\ln(7)x\amp=-\ln(7)-3\ln(5)\\ (\ln(5)-2\ln(7))x \amp=-\ln(7)-3\ln(5)\\ \frac{(\ln(5)-2\ln(7))x}{\color{red}{\ln(5)-2\ln(7)}} \amp=\frac{-\ln(7)-3\ln(5)}{\color{red}{\ln(5)-2\ln(7)}}\\ x \amp=\frac{-\ln(7)-3\ln(5)}{\ln(5)-2\ln(7)} \end{align*}

Notice that the answers we got look very different at first. However, they do actually work out to the same number, so they are both the correct answer. If you aren't sure, plug both of our answers into a calculator and check that it gives you the same decimal.

Checkpoint 17.11.

Solve the following equation:

\begin{equation*} 3^{x-7}=5^{4x} \end{equation*}
Answer.

The exact form of the answer will vary, depending on what log you took and how you simplified. Some examples of the correct answer include \(\frac{-7}{4\log_3(5)-1}\) and \(\frac{-7\ln(3)}{4\ln(5)-\ln(3)}\text{.}\) All of the answers will simplify to approximately \(-1.44036...\text{.}\)

Solution.

Unfortunately, since 3 is not a power of 5 and 5 is not a power of 7, there is no log we can pick that will cancel everything out. So, we have three options:

  1. Use log base 3: This would cancel the 3, but not the 5, so we would have a \(\log_3(5)\) floating around in our answer. If we are writing it on paper (like on a quiz or exam), that's no problem, but we can't enter bases for our logs into WeBWorK.

  2. Use log base 5: This would cancel the 5, but not the 3, so we would have a \(\log_5(3)\) floating around in our answer. If we are writing it on paper (like on a quiz or exam), that's no problem, but we can't enter bases for our logs into WeBWorK.

  3. Use natural log: This won't cancel the 5 or the 5, so we will have a \(\ln(3)\) and a \(\ln(5)\) in our answer. While that may be more to keep track of while solving the equation, the benefit is that we can type natural log into WeBWorK directly. The other benefit is that you don't have to think about it: you can just always use natural log, no matter what the bases are.

As you can see, there are benefits and drawbacks to each option. Any of them are totally fine, and even professional mathematicians have different preferences! Pick whichever makes the most sense to you. In this example, we will show both using log base 3 and natural log. The solution with log base 5 is very similar to the solution with log base 3, so we leave that to you to work out.

Option 1: Log base 3. We will start by taking the log base 3 of both sides, canceling what we can, and then using log rules to bring down the remaining exponents.

\begin{align*} {\color{red}{\log_3}}\left(3^{x-7}\right)\amp={\color{red}{\log_3}}\left(5^{4x}\right)\\ x-7\amp=(4x)\log_3(5) \end{align*}

This is where it becomes really important to look carefully at what we have. On one side we have a nice little \(x-7\text{.}\) On the other, we have this \(\log_3(5)\) floating around. Although it doesn't look like it right away, that's just a number. I could plug that into a calculator and it would spit out some kind of decimal. So, we will treat it like any other number.

\begin{align*} x-7\amp=(4x)\log_3(5)\\ x-7\amp=4\log_3(5)x\\ x-7{\color{red}{-x}}\amp=4\log_3(5)x{\color{red}{-x}}\\ -7\amp=4\log_3(5)x-x\\ -7\amp=(4\log_3(5)-1)x\\ \frac{-7}{\color{red}{4\log_3(5)-1}}\amp=\frac{(4\log_3(5)-1)x}{\color{red}{4\log_3(5)-1}}\\ \frac{-7}{4\log_3(5)-1}\amp=x \end{align*}

Option 3: Natural log. We will start by taking the natural log of both sides and using log rules to bring the exponents down.

\begin{align*} 3^{x-7}\amp=5^{4x}\\ {\color{red}{\ln}}\left(3^{x-7}\right)\amp={\color{red}{\ln}}\left(5^{4x}\right)\\ (x-7)\ln(3)\amp=(4x)\ln(5) \end{align*}

Again, we have to pay close attention here to what is a number. Although \(\ln(3)\) and \(\ln(5)\) don't immediately look like numbers, there are no variables in them. I can plug each of them into a calculator and it will spit out some ugly-looking decimal. So, I will treat each of them the same way I would any other number.

\begin{align*} (x-7)\ln(3)\amp=(4x)\ln(5)\\ \ln(3)x-7\ln(3)\amp=4\ln(5)x\\ \ln(3)x-7\ln(3){\color{red}{-\ln(3)x}}\amp=4\ln(5)x{\color{red}{-\ln(3)x}}\\ -7\ln(3)\amp= 4\ln(5)x-\ln(3)x\\ -7\ln(3) \amp=(4\ln(5)-\ln(3))x\\ \frac{-7\ln(3)}{\color{red}{4\ln(5)-\ln(3)}} \amp=\frac{(4\ln(5)-\ln(3))x}{\color{red}{4\ln(5)-\ln(3)}}\\ \frac{-7\ln(3)}{4\ln(5)-\ln(3)} \amp=x \end{align*}

Notice that the answers we got look very different at first. However, they do actually work out to the same number, so they are both the correct answer. If you aren't sure, plug both of our answers into a calculator and check that it gives you the same decimal.