3.3: Solving Exponential Equations

Because of the fact that logarithms are exponents, the rules for working with logarithms are similar to those that govern exponential expressions. One very helpful rule of equality for working with logarithms is related to the exponential rule for raising a power to a power. We recall one of the rules of exponents as:
$$\left(b^{x}\right)^{y}=b^{x * y}$$
in other words
$$\left(5^{2}\right)^{4}=\left(5^{2}\right)\left(5^{2}\right)\left(5^{2}\right)\left(5^{2}\right)=5^{2 * 4}=5^{8}$$
In logarithmic notation, this rule works out as:
$$\log _{b} M^{p}=p * \log _{b} M$$
The reason for this comes from the rule for exponents. Let's say that $\log _{b} M=x$
Then:
$$\log _{b} M=x$$
this means that
$$b^{x}=M$$
and
$$\begin{array}{c} \left(b^{x}\right)^{p}=(M)^{p} \\ \text { so } \\ b^{p * x}=M^{p} \end{array}$$

Now, we come back to the question of $\log _{b} M^{p}=?$. This expression $\left(\log _{b} M^{p}\right)$ is asking the question "What power do we raise $b$ to in order to get an answer of
$M^{p} ?$ The result on the previous page shows that:
$$b^{p x}=M^{p}$$
This means that we must raise $b$ to the $p x$ power to get an answer of $M^{p}$. Remember that $x=\log _{b} M .$ This means that:
$$\begin{array}{c} b^{p x}=M^{p} \\ \text { so } \\ \log _{b} M^{p}=p x=p * \log _{b} M \end{array}$$
This statement of equality is useful if we are trying to solve equations in which the variable is an exponent.
Example
Solve for $x$
$$4^{x}=53$$
We start by taking a logarithm on both of the equation. Just as we can add to both sides of an equation, or multiply on both sides of an equation, or raise both sides of an equation to a power, we can also take the logarithm of both sides. So long as two quantities are equal, then their logarithms will also be equal.
$$\begin{aligned} 4^{x} &=53 \\ \log 4^{x} &=\log 53 \\ x \log 4 &=\log 53 \\ x &=\frac{\log 53}{\log 4} \approx 2.864 \end{aligned}$$

since the log base 10 and log base $e$ are both programmed into most calculators, these are the most commonly used bases for logarithms.
Example
Solve for $x$
$$5^{2 x+3}=17$$
We start this problem in the same fashion, but this time we will use a logarithm to the base $e:$
$$\begin{array}{c} 5^{2 x+3}=17 \\ \ln 5^{2 x+3}=\ln 17 \\ (2 x+3) \ln 5=\ln 17 \end{array}$$
There are several possibilities for finishing the problem from this point. We will focus on two of them that are the most useful for solving more complex problems. First, we will distribute the $\ln 5$ into the parentheses and then get the $x$ by itself.

$$\begin{aligned} (2 x+3) \ln 5 &=\ln 17 \\ x * 2 \ln 5+3 \ln 5 &=\ln 17 \\ -3 \ln 5 &=-3 \ln 5 \\ x * 2 \ln 5 &=\ln 17-3 \ln 5 \\ x &=\frac{\ln 17-3 \ln 5}{2 \ln 5} \approx-0.620 \end{aligned}$$
And we can check the answer by plugging it back in:
$$5^{2 *(-0.0620)+3} \approx 5^{1.760} \approx 16.9897 \approx 17$$

We can also approximate the logarithms in the problem and solve for an approximate answer:
$$\begin{aligned} (2 x+3) \ln 5 &=\ln 17 \\ x * 2 \ln 5+3 \ln 5 &=\ln 17 \\ 3.2189 x+4.8283 & \approx 2.8332 \\ -4.8283 & \approx-4.8283 \\ 3.2189 x & \approx-1.9951 \\ x & \approx-0.620 \end{aligned}$$
If you use the method of approximating, it's important to make a good approximation. At least $4-5$ decimal places are necessary for an accurate answer.
Let's look at an example that has variables on both sides of the equation:

Example
Solve for $x$
$$4^{3 x}=9^{2 x-1}$$
We'll use log base 10 in this problem.
$$\begin{aligned} 4^{3 x} &=9^{2 x-1} \\ \log 4^{3 x} &=\log 9^{2 x-1} \\ 3 x * \log 4 &=(2 x-1) \log 9 \\ x * 3 \log 4 &=x * 2 \log 9-\log 9 \end{aligned}$$

If we collect like terms, we'll end up with:
$$\begin{aligned} x * 3 \log 4 &=x * 2 \log 9-\log 9 \\ \log 9 &=x * 2 \log 9-x * 3 \log 4 \end{aligned}$$
At this point, if we want to get the $x$ by itself, we need to factor out the $x$ on the right-hand side:
$$\begin{array}{l} \log 9=x * 2 \log 9-x * 3 \log 4 \\ \log 9=x(2 \log 9-3 \log 4) \end{array}$$

Then divide on both sides by the coefficient in parentheses:
$$\frac{\log 9}{2 \log 9-3 \log 4}=\frac{x\cancel{(2 \log 9-3 \log 4)}}{\cancel{2 \log 9-3 \log 4}} \\ \frac{\log 9}{2 \log 9-3 \log 4}=x \\ 9.327 \approx x$$
Again, we can check our answer by plugging it back into the equation:
$$4^{3 * 9.327} \approx 4^{27.981} \approx 7.0184 * 10^{16}$$
$$9^{2 * 9.327-1} \approx 9^{17.654} \approx 7.0177 * 10^{16}$$

We could also have solved this equation by approximating the logarithms in the beginning.
$$\begin{aligned} 4^{3 x} &=9^{2 x-1} \\ \log 4^{3 x} &=\log 9^{2 x-1} \\ 3 x * \log 4 &=(2 x-1) \log 9 \\ 3 x(0.60206) & \approx(2 x-1) 0.95424 \\ 1.80618 x & \approx 1.9085 x-0.95424 \\ 0.95424 & \approx 0.10232 x \\ 9.326 & \approx x \end{aligned}$$
This answer is less accurate than the other approximation $(9.326036 \text { vs. } 9.327424)$ The accuracy of an answer depends upon the original approximations for the logarithms.

Solve for the indicated variable.
1) $\quad 2^{x}=5$
2) $\quad 2^{x}=9$
3) $\quad 3^{x}=7$
4) $\quad 3^{x}=20$
5) $\quad 2^{x+1}=6$
6) $\quad 7^{x+1}=41$
7) $\quad 5^{x+1}=36$
8) $\quad 8^{x-2}=6$
9) $\quad 4^{2 x+3}=50$
10) $\quad 4^{x+2}=5^{x}$
11) $\quad 5^{2 x+1}=9$
12) $\quad 6^{x+4}=10^{x}$
13) $\quad 7^{y+1}=3^{y}$
14) $\quad 2^{x+1}=3^{x-2}$
15) $\quad 6^{y+2}=5^{y}$
16) $\quad 7^{x-3}=3^{x+1}$
17) $\quad 6^{2 x+1}=5^{x+2}$
18) $\quad 9^{1-x}=12^{x+1}$
19) $\quad 5^{2 x-1}=3^{x-3}$
20) $\quad 3^{x-2}=4^{2 x+1}$
21) $\quad 8^{3 x-2}=9^{x+2}$
22) $\quad 2^{2 x-3}=5^{-x-1}$
23) $\quad 10^{3 x+2}=5^{x+3}$
24) $\quad 5^{3 x}=3^{x+4}$
25) $\quad 3^{x+4}=2^{1-3 x}$
26) $\quad 4^{2 x+3}=5^{x-2}$
27) $\quad 3^{2-3 x}=4^{2 x+1}$
28) $\quad 2^{2 x-3}=5^{x-2}$