Exponential equations are equations where the unknown is in the exponent. For example:
$3^x=81$3x=81
$5\times2^x=40$5×2x=40
$7^{5x-2}=20$75x−2=20
To solve this type of equation without technology of the use of logarithms, we need to write both sides of the equation with the same base, then we can equate the indices.
When both sides of an exponential equation are written with the same base, we can equate the indices:
If $a^x=a^y$ax=ay, then $x=y$x=y.
To write both sides with the same base it is helpful to be familiar with low powers of prime numbers, so you can recognise them and rewrite them in index form. It is also vital to be confident with the use of index laws.
Worked examples
Example 1
Solve $5^p=125$5p=125, for $p$p.
Think: One side of the equation is a power of $5$5, can we write the other side also as a power of $5$5?
Do:
| $5^p$5p | $=$= | $125$125 | |
| $5^p$5p | $=$= | $5^3$53 | |
| $\therefore$∴ $p$p | $=$= | $3$3 | , by equating indices |
Example 2
Solve $\left(27\right)^{x+1}=\frac{1}{81}$(27)x+1=181, for $x$x.
Think: Can you spot a common base that both sides could be written in? Both $27$27 and $81$81 are powers of $3$3.
Do: Use index laws to write both sides as a single power of three 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 3
Solve $5\times16^y=40\times\sqrt[3]{32}$5×16y=40×3√32, for $y$y.
Think: We have a power of $2$2 on the left-hand side but it is multiplied by a $5$5. If we first divide both sides by the factor of $5$5, can we then write both sides as powers of $2$2?
Do:
| $5\times16^y$5×16y | $=$= | $40\times\sqrt[3]{32}$40×3√32 |
| $\left(16\right)^y$(16)y | $=$= | $8\times\sqrt[3]{32}$8×3√32 |
We now have an equation with $16$16, $8$8 and $32$32 which can all be written as powers of $2$2. Proceed with index laws and remember $\sqrt[n]{x}=x^{\frac{1}{n}}$n√x=x1n.
| $\left(16\right)^y$(16)y | $=$= | $8\times\sqrt[3]{32}$8×3√32 | |
| $\left(2^4\right)^y$(24)y | $=$= | $2^3\times\left(2^5\right)^{\frac{1}{3}}$23×(25)13 | |
| $2^{4y}$24y | $=$= | $2^{\left(3+\frac{5}{3}\right)}$2(3+53) | |
| $2^{4y}$24y | $=$= | $2^{\frac{14}{3}}$2143 | |
| Hence, $4y$4y | $=$= | $\frac{14}{3}$143 | , by equating indices |
| $\therefore y$∴y | $=$= | $\frac{14}{12}$1412 |
Example 4
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.
Question 3
Consider the equation
$\left(2^x\right)^2-9\times2^x+8=0$(2x)2−9×2x+8=0
The equation can be reduced to a quadratic equation by using a certain substitution.
By filling in the gaps, determine the correct substitution that would reduce the equation to a quadratic.
Let $m=\left(\editable{}\right)^{\editable{}}$m=()
Solve the equation for $x$x by using the substitution $m=2^x$m=2x.