Use the definition of a logarithm to solve logarithmic equations

We have already seen that every logarithmic equation [latex]{\mathrm{log}}_{b}\left(x\right)=y[/latex] is equivalent 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 in compact form and then apply the definition of logs to solve for x:

[latex]\begin{cases}{\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{Apply the definition of a logarithm}.\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 by 6}.\hfill \end{cases}[/latex]

A General Note: Using the Definition of a Logarithm to Solve Logarithmic Equations

For any algebraic expression S and real numbers b and c, 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]

Example 9: Using Algebra to Solve a Logarithmic Equation

Solve [latex]2\mathrm{ln}x+3=7[/latex].


[latex]\begin{cases}2\mathrm{ln}x+3=7\hfill & \hfill \\ \text{ }2\mathrm{ln}x=4\hfill & \text{Subtract 3}.\hfill \\ \text{ }\mathrm{ln}x=2\hfill & \text{Divide by 2}.\hfill \\ \text{ }x={e}^{2}\hfill & \text{Rewrite in exponential form}.\hfill \end{cases}[/latex]

Try It 9

Solve [latex]6+\mathrm{ln}x=10[/latex].


Example 10: Using Algebra Before and After Using the Definition of the Natural Logarithm

Solve [latex]2\mathrm{ln}\left(6x\right)=7[/latex].


[latex]\begin{cases}2\mathrm{ln}\left(6x\right)=7\hfill & \hfill \\ \text{ }\mathrm{ln}\left(6x\right)=\frac{7}{2}\hfill & \text{Divide 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 by 6}.\hfill \end{cases}[/latex]

Try It 10

Solve [latex]2\mathrm{ln}\left(x+1\right)=10[/latex].


Example 11: Using a Graph to Understand the Solution to a Logarithmic Equation

Solve [latex]\mathrm{ln}x=3[/latex].


[latex]\begin{cases}\mathrm{ln}x=3\hfill & \hfill \\ x={e}^{3}\hfill & \text{Use the definition of the natural logarithm}\text{.}\hfill \end{cases}[/latex]

Figure 2 represents the graph of the equation. On the graph, the x-coordinate of the point at which 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].

Graph of two questions, y=3 and y=ln(x), which intersect at the point (e^3, 3) which is approximately (20.0855, 3).
Figure 2. 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).

Try It 11

Use a graphing calculator to estimate the approximate solution to the logarithmic equation [latex]{2}^{x}=1000[/latex] to 2 decimal places.