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 [latex]x > 0[/latex], [latex]S > 0[/latex], [latex]T > 0[/latex] and any positive real number [latex]b[/latex], where [latex]b\ne 1[/latex],
[latex]{\mathrm{log}}_{b}S={\mathrm{log}}_{b}T\text{ if and only if }S=T[/latex]
[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 [latex]x[/latex]and we get [latex]x = 9[/latex]. To check, we can substitute [latex]x = 9[/latex] 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.
[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 [latex]x = 10[/latex] 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]
one-to-one property of logarithmic functions
For any real numbers [latex]x > 0[/latex], [latex]S > 0[/latex], [latex]T > 0[/latex] and any positive real number [latex]b[/latex], where [latex]b\ne 1[/latex],
[latex]{\mathrm{log}}_{b}S={\mathrm{log}}_{b}T\text{ if and only if }S=T[/latex]
Note: When solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.
- Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation is of the form [latex]{\mathrm{log}}_{b}S={\mathrm{log}}_{b}T[/latex].
- Use the one-to-one property to set the arguments equal to each other.
- Solve the resulting equation, [latex]S = T[/latex], for the unknown.