11.08 Equations using logarithms (Enrichment)
We previously saw that a logarithm is the index that an exponential is raised to in order to get a particular result. In other words, logarithms are the inverse functions of exponentials, and exponentials are the inverse functions of logarithms.
This gives us a method to solve any equation with exponentials or logarithms. Following the general method for solving equations we perform the reverse operation to isolate the variable. When the variable is inside an exponential function we take the logarithm of the exponential, and when the variable is in a logarithmic function we take the exponential of the logarithm.
Worked examples
Example 1
Solve $\log_5\left(4x-11\right)-9=-7$log5(4x−11)−9=−7 for $x$x.
Think: We will rearrange the equation to make $x$x the subject. To reverse the logarithm of base $5$5, we can take $5$5 to the power of both sides of the equation.
Do:
| $\log_5\left(4x-11\right)-9$log5(4x−11)−9 | $=$= | $-7$−7 |
|
| $\log_5\left(4x-11\right)$log5(4x−11) | $=$= | $2$2 |
Adding $9$9 to both sides of the equation |
| $5^{\log_5\left(4x-11\right)}$5log5(4x−11) | $=$= | $5^2$52 |
Taking $5$5 to the power of each side of the equation |
| $4x-11$4x−11 | $=$= | $25$25 |
Simplifying both sides of the equation. We have used the rule $B^{\log_Bn}=n$BlogBn=n to simplify the left hand side. |
| $4x$4x | $=$= | $36$36 |
Adding $11$11 to both sides of the equation. |
| $x$x | $=$= | $9$9 |
Dividing both sides of the equation by $4$4. |
Reflect: We used the rule $B^{\log_Bn}=n$BlogBn=n which is a property of inverse functions. In fact, each step we took involved applying an inverse function. For example, if $f\left(x\right)=x-9$f(x)=x−9 then $f^{-1}\left(x\right)=x+9$f−1(x)=x+9, so we cancelled out the $-9$−9 by applying the inverse function.
Example 2
Solve $5\left(8^{\frac{x}{12}}\right)=640$5(8x12)=640 for $x$x.
Think: We will rearrange the equation to make $x$x the subject. To reverse the exponential of base $8$8, we can take the logarithm of base $10$10 on both sides of the equation and use the rule $\log A^n=n\log A$logAn=nlogA.
Do:
| $5\left(8^{\frac{x}{12}}\right)$5(8x12) | $=$= | $640$640 |
|
| $8^{\frac{x}{12}}$8x12 | $=$= | $128$128 |
Dividing both sides of the equation by $5$5 |
| $\log8^{\frac{x}{12}}$log8x12 | $=$= | $\log128$log128 |
Taking the logarithm base $10$10 of both sides of the equation |
| $\frac{x}{12}\log8$x12log8 | $=$= | $\log128$log128 |
Using the rule $\log A^n=n\log A$logAn=nlogA |
| $x$x | $=$= | $\frac{12\log128}{\log8}$12log128log8 |
Dividing both sides by $\frac{\log8}{12}$log812 |
| $x$x | $=$= | $28$28 |
Evaluating the right hand side |
Reflect: We used the rule $\log A^n=n\log A$logAn=nlogA which is also a consequence of inverse functions. The reason we chose base $10$10 was to make it easier to calculate the solution using a calculator. In other cases it might be better to use a logarithm with the same base as the exponential so that the solution is a logarithmic expression.
Exponentials and logarithms of the same base are inverse functions of one another.
This fact allows us to solve equations involving exponentials and logarithms by applying the inverse function to both sides in order to isolate the variable.
Practice questions
Question 1
The graphs of $y=3^x$y=3x (labelled $B$B) and $y=x$y=x (labelled $A$A) have been plotted below.
By reflecting $y=3^x$y=3x about the line $y=x$y=x, plot the graph of the inverse of $y=3^x$y=3x.
- Loading Graph...
Question 2
Solve $\log_{64}x=\frac{1}{3}$log64x=13 for $x$x.
Question 3
Solve $3\left(10^x\right)=6$3(10x)=6 for the exact value of $x$x.