Exponential functions have many applications in growth and decay such as population growth, investment growth, radioactive decay and depreciation. Before looking at some applications in further detail, let's review how to solve some exponential equations.
Exponential equations
We want to look at solving problems where the unknown is in the exponent–these are called exponential equations. They look like:
$3^x=81$3x=81, $5\times2^x=40$5×2x=40 and $7^{5x-2}=20$75x−2=20
In some cases algebraic manipulation using our index laws will allow us to solve exponential equations without a calculator. To achieve this we need to write both sides of the equation with the same base, then we can equate the indices. If we cannot write both sides of the equation with the same base we could use technology to solve these or we can use logarithms, which we will explore in further detail later in this course.
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 index laws.
Worked examples
Example 1
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 |
Writing down the equation |
| $\left(3^3\right)^{x+1}$(33)x+1 | $=$= | $\frac{1}{3^4}$134 |
Expressing both sides using powers of $3$3 |
| $3^{3x+3}$33x+3 | $=$= | $3^{-4}$3−4 |
Using the index law $A^{-n}=\frac{1}{A^n}$A−n=1An |
| Hence $3x+3$3x+3 | $=$= | $-4$−4 |
Equating exponents |
| $3x$3x | $=$= | $-7$−7 |
Moving the constant terms to one side |
| $\therefore$∴ $x$x | $=$= | $-\frac{7}{3}$−73 |
Solving for $x$x |
Example 2
Solve $5\times16^y=40\times\sqrt[3]{32}$5×16y=40×3√32, for $y$y.
Think: On the left-hand side, we have a power of $2$2, but it's 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 |
Writing down the equation |
| $\left(2^4\right)^y$(24)y | $=$= | $2^3\times\left(2^5\right)^{\frac{1}{3}}$23×(25)13 |
Expressing both sides using powers of $2$2 |
| $2^{4y}$24y | $=$= | $2^{\left(3+\frac{5}{3}\right)}$2(3+53) |
Using index laws |
| $2^{4y}$24y | $=$= | $2^{\frac{14}{3}}$2143 |
Combining terms in the exponent |
| Hence $4y$4y | $=$= | $\frac{14}{3}$143 |
Equating exponents |
| $y$y | $=$= | $\frac{14}{12}$1412 |
Solving for $y$y
|
| $y$y | $=$= | $\frac{7}{6}$76 |
Simplifying |
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 $8^{x+5}=\frac{1}{32\sqrt{2}}$8x+5=132√2.