If an exponential equation cannot be rearranged so that both sides have the same base, then there are two methods to solve such equations of the form $a^x=b$ax=b:
Method 1: Take the logarithm of both sides, or
Method 2: Rearrange the equation into logarithmic form and use change of base formula.
Both methods will let us rearrange the equation to get $x$x in terms of $\log a$loga and $\log b$logb. However, it is often easier to use method 1 of taking the logarithm of both sides.
Solve $12^x=30$12x=30 for $x$x as both an exact value and to three decimal places.
Think: This is an equation of the form $a^x=b$ax=b. So we can solve it by taking the logarithm of both sides (method 1). This might seem like a strange thing to do, but remember, as long as we do the same thing to both sides of an equation, it will continue to be true!
Do:
$12^x$12x | $=$= | $30$30 | |
$\log12^x$log12x | $=$= | $\log30$log30 |
Take the logarithm of both sides |
$x\log12$xlog12 | $=$= | $\log30$log30 |
Using the identity $\log A^B=B\log A$logAB=BlogA |
$x$x | $=$= | $\frac{\log30}{\log12}$log30log12 |
|
$\approx$≈ | $1.369$1.369 |
Using a calculator, and rounding to three decimal places |
Reflect: When taking the log of both sides, it is easier to use logarithms of base $10$10 or $e$e, in order to evaluate the solution with a calculator.
For comparison, here is how we solve the same equation by rewriting it in logarithmic form (method 2):
$12^x$12x | $=$= | $30$30 | |
$x$x | $=$= | $\log_{12}30$log1230 |
Using the relationship between exponential and logarithms |
$x$x | $=$= | $\frac{\log_{10}30}{\log_{10}12}$log1030log1012 |
Using the change of base law |
$\approx$≈ | $1.369$1.369 |
Rounding to three decimal places |
For exponential equations in the form $a^x=b$ax=b, there is exists a solution only if $b>0$b>0!
For example, $4^x=-1$4x=−1 and $7^{\left(x+3\right)}=-49$7(x+3)=−49 do not have any solutions, since we cannot raise the numbers $4$4 or $7$7 to any power and get a negative number.
Which of the following is a good estimate for the value of $x$x, if $3^x=29$3x=29?
$-4
$3
$2
$4
Consider the equation $2^x=11$2x=11.
Rearrange the equation into the form $x=\frac{\log A}{\log B}$x=logAlogB.
Evaluate $x$x to three decimal places.
Consider the equation $2^{9x-4}=90$29x−4=90.
Make $x$x the subject of the equation.
Evaluate $x$x to three decimal places.