3.4: Solving Logarithmic Equations

In the previous section, we took exponential equations and used the properties of logarithms to restate them as logarithmic equations. In this section, we will take logarithmic equations and use properties of logarithms to restate them as exponential equations. In the previous section, we used the property of logarithms that said $\log _{b} M^{p}=p \log _{b} M .$ In this section, we will make use of two additional properties of logarithms:
$$\log _{b}(M * N)=\log _{b} M+\log _{b} N$$
and
$$\log _{b} \frac{M}{N}=\log _{b} M-\log _{b} N$$
Just as our previous property of logarithms was simply a restatement of the rules of expoenents, these two properties of logarithms depend on the rules of exponents as well. since we're interested in $\log _{b} M$ and $\log _{b} N,$ let's restate these in terms of exponents:
If $\log _{b} M=x$ then $b^{x}=M$ and if $\log _{b} N=y$ then $b^{y}=N$
The properties of logarithms we're interested in justifying have to do with $M * N$ and $\frac{M}{N},$ so let's look at those expressions in terms of exponents:
$$\begin{array}{c} M * N=b^{x} * b^{y}=b^{x+y} \\ \text { and } \\ \frac{M}{N}=\frac{b^{x}}{b^{y}}=b^{x-y} \end{array}$$

If we're interested in $\log _{b}(M * N),$ then we're asking the question "What power do we raise $b$ to in order to get $M * N ? "$ We can see above that raising $b$ to the $x+y$ power gives us $M * N .$ since $x=\log _{b} M$ and $y=\log _{b} N$ then $x+y=$ $\log _{b} M+\log _{b} N,$ so:
$$\log _{b}(M * N)=x+y=\log _{b} M+\log _{b} N$$
Likewise, if we're interested in $\log _{b} \frac{M}{N},$ we're asking the question "What power do we raise $b$ to in order to get $\frac{M}{N} ? "$ since raising $b$ to the $x-y$ power gives us $\frac{M}{N}$ and $x-y=\log _{b} M-\log _{b} N,$ then:
$$\log _{b} \frac{M}{N}=x-y=\log _{b} M-\log _{b} N$$

Let's look at an example to see how we'll use this to solve equations:

Example
Solve for $x$
$$\log _{2} x+\log _{2}(x-4)=2$$
The first thing we can do here is to combine the two logarithmic statements into one. since $\log _{b}(M * N)=\log _{b} M+\log _{b} N,$ then $\log _{2} x+\log _{2}(x-4)=\log _{2}[x(x-4)]$
$$\begin{aligned} \log _{2} x+\log _{2}(x-4) &=2 \\ \log _{2}[x(x-4)] &=2 \end{aligned}$$
Then we'll restate the resulting logarithmic relationship as an exponential relationship:
$$\begin{aligned} 2^{2} &=x(x-4) \\ 4 &=x^{2}-4 x \\ 0 &=x^{2}-4 x-4 \\ 4.828,-0.828 & \approx x \end{aligned}$$

Most textbooks reject answers that result in taking the logarithm of a negative number, such as would be the case for $x \approx-0.828 .$ However, the logarithms of negative numbers result in complex valued answers, rather than an undefined quantity. For that reason, in this text, we will include all answers.

If a problem involves a difference of logarithms, we can use the other property of logarithms introduced in this section.
Example
Solve for $x$
$$\log (5 x-1)-\log (x-2)=2$$
Again, our first step is to restate the difference of logarithms using the property $\log _{b} \frac{M}{N}=\log _{b} M-\log _{b} N:$
$$\begin{aligned} \log (5 x-1) &-\log (x-2)=2 \\ \log \left[\frac{5 x-1}{x-2}\right] &=2 \end{aligned}$$
We're working with a logarithm in base 10 in this problem, so in our next step we'll say:
$$\begin{array}{c} \log \left[\frac{5 x-1}{x-2}\right]=2 \\ \frac{5 x-1}{x-2}=10^{2} \end{array}$$

Then multiply on both sides by $x-2$
$$\begin{array}{c} 10^{2}=\frac{5 x-1}{x-2} \\ (x-2) * 100=\frac{5 x-1}{\cancel{(x-2)}} *\cancel{(x-2)} \end{array}$$

And, solve for $x$
$$\begin{array}{c} 100 x-200=5 x-1 \\ 95 x=199 \\ x=\frac{199}{95} \end{array}$$
In some equations, all of the terms are stated using logartihms. These equations often come out in a form that says $\log _{b} x=\log _{b} y .$ If this is the case, we can then conclude that $x=y$

It seems reasonable that if the exponent we raise $b$ to in order to get $x$ is the same exponent that we raise $b$ to in order to get $y,$ then $x$ and $y$ are the same thing.
Assume:
$$\log _{b} x=\log _{b} y$$
then
$$b^{a}=x \text { and } b^{a}=y$$
if both $x$ and $y$ are equal to $b^{a},$ then $x=y$

Example
Solve for $x$
$\log _{5}(4-x)=\log _{5}(x+8)+\log _{5}(2 x+13)$

First, let's use the properties of logarithms to restate the equation so that there is only one logarithm on each side.
$$\begin{array}{l} \log _{5}(4-x)=\log _{5}(x+8)+\log _{5}(2 x+13) \\ \log _{5}(4-x)=\log _{5}[(x+8)(2 x+13)] \end{array}$$
Then, we'll use the property of logarithms we just discussed:
$$\text { If } \log _{b} x=\log _{b} y$$
then
$$\begin{array}{c} x=y \\ \log _{5}(4-x)=\log _{5}[(x+8)(2 x+13)] \\ 4-x=(x+8)(2 x+13) \\ 0=2 x^{2}+29 x+104 \\ 0=2(x+5)(x+10) \\ -5,-10=x \end{array}$$

Exercises 3.4

Solve for the indicated variable in each equation.
1) $\quad \log _{3} 5+\log _{3} x=2$
2) $\quad \log _{4} x+\log _{4} 5=1$
3) $\quad \log _{2} x=2+\log _{2} 3$
4) $\quad \log _{5} x=2+\log _{5} 3$
5) $\quad \log _{3} x+\log _{3}(x-8)=2$
6) $\quad \log _{6} x+\log _{6}(x-5)=1$
7) $\quad \log (3 x+2)=\log (x-4)+1$
8) $\quad \log (x-1)-\log x=-0.5$
9) $\quad \log _{2} a+\log _{2}(a+2)=3$
10) $\quad \log _{3} x+\log _{3}(x-2)=1$
11) $\quad \log _{2} y-\log _{2}(y-2)=3$
12) $\quad \log _{2} x-\log _{2}(x+3)=2$
13) $\quad \log _{3} x+\log _{3}(x+4)=2$
14) $\quad \log _{4} u+\log _{4}(u+1)=1$
15) $\quad \log 5+\log x=\log 6$
16) $\quad \ln x+\ln 4=\ln 2$
17) $\quad \log _{7} x-\log _{7} 12=\log _{7} 2$
18) $\quad \log 2-\log x=\log 8$
19) $\quad \log _{3} x-\log _{3}(x-2)=\log _{3} 4$
20) $\quad \log _{6} 2-\log _{6}(x-2)=\log _{6} 9$
21) $\quad \log _{4} x-\log _{4}(x-4)=\log _{4}(x-6)$
22) $\quad \log _{9}(2 x+7)-\log _{9}(x-1)=\log _{9}(x-7)$
23) $\quad 2 \log _{2} x=\log _{2}(2 x-1)$
24) $\quad 2 \log _{4} y=\log _{4}(y+2)$
25) $\quad 2 \log (x-3)-3 \log 2=1$
26) $\quad 2 \log _{5} 7-\log _{5}(x+1)=\log _{5}(2 x-5)$