Logarithmic Equations

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]

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

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

Show Solution

[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 x-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).

Using the One-to-One Property of Logarithms to Solve Logarithmic Equations

As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers x > 0, S > 0, T > 0 and any positive real number b, where [latex]b\ne 1[/latex],

[latex]{\mathrm{log}}_{b}S={\mathrm{log}}_{b}T\text{ if and only if }S=T[/latex]

For example,

[latex]\text{If }{\mathrm{log}}_{2}\left(x - 1\right)={\mathrm{log}}_{2}\left(8\right),\text{then }x - 1=8[/latex]

So if [latex]x - 1=8[/latex], then we can solve for x and we get x = 9. To check, we can substitute x = 9 into the original equation: [latex]{\mathrm{log}}_{2}\left(9 - 1\right)={\mathrm{log}}_{2}\left(8\right)=3[/latex]. In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.

For example, consider the equation [latex]\mathrm{log}\left(3x - 2\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(x+4\right)[/latex]. To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm and then apply the one-to-one property to solve for x:

[latex]\begin{array}{l}\mathrm{log}\left(3x - 2\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(x+4\right)\hfill & \hfill \\ \text{}\mathrm{log}\left(\frac{3x - 2}{2}\right)=\mathrm{log}\left(x+4\right)\hfill & \text{Apply the quotient rule of logarithms}.\hfill \\ \text{}\frac{3x - 2}{2}=x+4\hfill & \text{Apply the one-to-one property}.\hfill \\ \text{}3x - 2=2x+8\hfill & \text{Multiply both sides of the equation by }2.\hfill \\ \text{}x=10\hfill & \text{Subtract 2}x\text{ and add 2}.\hfill \end{array}[/latex]

To check the result, substitute x = 10 into [latex]\mathrm{log}\left(3x - 2\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(x+4\right)[/latex].

[latex]\begin{array}{l}\mathrm{log}\left(3\left(10\right)-2\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(\left(10\right)+4\right)\hfill & \hfill \\ \text{}\mathrm{log}\left(28\right)-\mathrm{log}\left(2\right)=\mathrm{log}\left(14\right)\hfill & \hfill \\ \text{}\mathrm{log}\left(\frac{28}{2}\right)=\mathrm{log}\left(14\right)\hfill & \text{The solution checks}.\hfill \end{array}[/latex]