We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

Study Guides > College Algebra

Use the definition of a logarithm to solve logarithmic equations

We have already seen that every logarithmic equation logb(x)=y{\mathrm{log}}_{b}\left(x\right)=y is equivalent to the exponential equation by=x{b}^{y}=x. 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 log2(2)+log2(3x5)=3{\mathrm{log}}_{2}\left(2\right)+{\mathrm{log}}_{2}\left(3x - 5\right)=3. 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:

{log2(2)+log2(3x5)=3 log2(2(3x5))=3Apply the product rule of logarithms. log2(6x10)=3Distribute. 23=6x10Apply the definition of a logarithm. 8=6x10Calculate 23. 18=6xAdd 10 to both sides. x=3Divide by 6.\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}

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 b>0, b1b>0,\text{ }b\ne 1,

logb(S)=cif and only ifbc=S{\mathrm{log}}_{b}\left(S\right)=c\text{if and only if}{b}^{c}=S

Example 9: Using Algebra to Solve a Logarithmic Equation

Solve 2lnx+3=72\mathrm{ln}x+3=7.

Solution

{2lnx+3=7 2lnx=4Subtract 3. lnx=2Divide by 2. x=e2Rewrite in exponential form.\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}

Try It 9

Solve 6+lnx=106+\mathrm{ln}x=10.

Solution

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

Solve 2ln(6x)=72\mathrm{ln}\left(6x\right)=7.

Solution

{2ln(6x)=7 ln(6x)=72Divide by 2. 6x=e(72)Use the definition of ln. x=16e(72)Divide by 6.\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}

Try It 10

Solve 2ln(x+1)=102\mathrm{ln}\left(x+1\right)=10.

Solution

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

Solve lnx=3\mathrm{ln}x=3.

Solution

{lnx=3x=e3Use the definition of the natural logarithm.\begin{cases}\mathrm{ln}x=3\hfill & \hfill \\ x={e}^{3}\hfill & \text{Use the definition of the natural logarithm}\text{.}\hfill \end{cases}

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 e320{e}^{3}\approx 20. A calculator gives a better approximation: e320.0855{e}^{3}\approx 20.0855.

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 y=lnxy=\mathrm{ln}x and = 3 cross at the point (e3,3)\left(e^3,3\right), which is approximately (20.0855, 3).

Try It 11

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

Solution

Licenses & Attributions