We have already seen that every logarithmic equation [latex]{\mathrm{log}}_{b}\left(x\right)=y[/latex] is equal to the exponential equation [latex]{b}^{y}=x[/latex]. We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.
For example, consider the equation [latex]{\mathrm{log}}_{2}\left(2\right)+{\mathrm{log}}_{2}\left(3x - 5\right)=3[/latex]. To solve this equation, we can use rules of logarithms to rewrite the left side as a single log and then apply the definition of logs to solve for [latex]x[/latex]:
[latex]\begin{array}{l}{\mathrm{log}}_{2}\left(2\right)+{\mathrm{log}}_{2}\left(3x - 5\right)=3\hfill & \hfill \\ \text{ }{\mathrm{log}}_{2}\left(2\left(3x - 5\right)\right)=3\hfill & \text{Apply the product rule of logarithms}.\hfill \\ \text{ }{\mathrm{log}}_{2}\left(6x - 10\right)=3\hfill & \text{Distribute}.\hfill \\ \text{ }{2}^{3}=6x - 10\hfill & \text{Convert to exponential form}.\hfill \\ \text{ }8=6x - 10\hfill & \text{Calculate }{2}^{3}.\hfill \\ \text{ }18=6x\hfill & \text{Add 10 to both sides}.\hfill \\ \text{ }x=3\hfill & \text{Divide both sides by 6}.\hfill \end{array}[/latex]
using the definition of a logarithm to solve logarithmic equations
For any algebraic expression [latex]S[/latex] and real numbers [latex]b[/latex] and [latex]c[/latex], where [latex]b>0,\text{ }b\ne 1[/latex],
[latex]{\mathrm{log}}_{b}\left(S\right)=c\text{ if and only if }{b}^{c}=S[/latex]
Solve [latex]2\mathrm{ln}x+3=7[/latex].
[latex]\begin{array}{l}2\mathrm{ln}x+3=7\hfill & \hfill \\ \text{}2\mathrm{ln}x=4\hfill & \text{Subtract 3 from both sides}.\hfill \\ \text{}\mathrm{ln}x=2\hfill & \text{Divide both sides by 2}.\hfill \\ \text{}x={e}^{2}\hfill & \text{Rewrite in exponential form}.\hfill \end{array}[/latex]
As was the case when using the properties and rules of exponents and logarithms to rewrite expressions containing them, there can be more than one good way to solve a logarithmic equation. It is good practice to follow the examples given for each of the situations in this section, but you should think about alternative ways to creatively and correctly apply the properties and rules.Solve [latex]2\mathrm{ln}\left(6x\right)=7[/latex].
[latex]\begin{array}{l}2\mathrm{ln}\left(6x\right)=7\hfill & \hfill \\ \text{}\mathrm{ln}\left(6x\right)=\frac{7}{2}\hfill & \text{Divide both sides by 2}.\hfill \\ \text{}6x={e}^{\left(\frac{7}{2}\right)}\hfill & \text{Use the definition of }\mathrm{ln}.\hfill \\ \text{}x=\frac{1}{6}{e}^{\left(\frac{7}{2}\right)}\hfill & \text{Divide both sides by 6}.\hfill \end{array}[/latex]
Solve [latex]\mathrm{ln}x=3[/latex].
[latex]\begin{array}{l}\mathrm{ln}x=3\hfill & \hfill \\ x={e}^{3}\hfill & \text{Use the definition of }\mathrm{ln}\text{.}\hfill \end{array}[/latex]
Below is a graph of the equation. On the graph the [latex]x[/latex]-coordinate of the point where the two graphs intersect is close to 20. In other words [latex]{e}^{3}\approx 20[/latex]. A calculator gives a better approximation: [latex]{e}^{3}\approx 20.0855[/latex].
The graphs of [latex]y=\mathrm{ln}x[/latex] and y = 3 cross at the point [latex]\left(e^3,3\right)[/latex] which is approximately (20.0855, 3).
