6.04 Solving exponential equations
Solving by equating exponents
We now want to look at solving problems where the unknown is in the index, or exponent. These are exponential equations such as $3^x=81$3x=81, $5\times2^x=40$5×2x=40 and $7^{5x-2}=20$75x−2=20.
We could use technology to solve these, potentially involving logarithms when we can't easily find integer solutions. In many cases, however, algebraic manipulation using index laws will allow us to solve exponential equations without a calculator. To achieve this we want to write both sides of the equation with the same base, so that we can then equate the indices.
When both sides of an exponential equation are written with the same base, we can equate the indices. That is,
if $a^x=a^y$ax=ay, then $x=y$x=y.
Worked examples
example 1
Solve $\left(27\right)^{x+1}=\frac{1}{81}$(27)x+1=181, for $x$x.
Think: Both $27$27 and $81$81 are powers of $3$3, so let's rewrite the exponentials using $3$3 as our common base.
Do: Use index laws to write both sides as a single power of $3$3 and then equate the indices.
| $\left(27\right)^{x+1}$(27)x+1 | $=$= | $\frac{1}{81}$181 | |
| $\left(3^3\right)^{x+1}$(33)x+1 | $=$= | $\frac{1}{3^4}$134 | |
| $3^{3x+3}$33x+3 | $=$= | $3^{-4}$3−4 | |
| Hence. $3x+3$3x+3 | $=$= | $-4$−4 | , by equating indices |
| $3x$3x | $=$= | $-7$−7 | |
| $\therefore$∴ $x$x | $=$= | $-\frac{7}{3}$−73 |
Example 2
Solve $2^{2x}-20\left(2^x\right)+64=0$22x−20(2x)+64=0 for $x$x.
Think: We have index rules for multiplying and dividing powers but not adding. Since we have three terms added together here we will need a different starting approach. Notice the first term can be written as $\left(2^x\right)^2$(2x)2 and the middle term also has a $2^x$2x. In fact, we have a quadratic equation in terms of $2^x$2x. This may be easier to see if we make a substitution.
Do: Let $y=2^x$y=2x, then:
| $2^{2x}-20\left(2^x\right)+64$22x−20(2x)+64 | $=$= | $0$0 |
| $\left(2^x\right)^2-20\left(2^x\right)+64$(2x)2−20(2x)+64 | $=$= | $0$0 |
| Make the substitution $y$y | $=$= | $2^x$2x |
| $y^2-20y+64$y2−20y+64 | $=$= | $0$0 |
| $\left(y-16\right)\left(y-4\right)$(y−16)(y−4) | $=$= | $0$0 |
| $\therefore$∴ $y$y | $=$= | $4$4 or $16$16 |
Since $y=2^x$y=2x, we have $2^x=4$2x=4 or $2^x=16$2x=16, and thus our solutions are $x=2$x=2 or $x=4$x=4.
Practice questions
Question 1
Solve the equation $\left(2^2\right)^{x+7}=2^3$(22)x+7=23 for $x$x.
Question 2
Solve the equation $25^{x+1}=125^{3x-4}$25x+1=1253x−4.
Solving by using logarithms
Two methods to solve equations of the form $a^x=b$ax=b is to:
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.
Worked example
Example 3
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.
Practice questions
Question 3
Which of the following is a good estimate for the value of $x$x, if $3^x=29$3x=29?
$-4
A$3
B$2
C$4
D
Question 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.
Question 5
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.