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1.025 Exponential equations

Lesson

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 look at 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$75x2=20

We could use technology to solve these or algebraically using logarithms, which we will explore in further detail later in this course. However, in many 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.

Equating 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 our 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}$34

Using the index law $A^{-n}=\frac{1}{A^n}$An=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=40×332, 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}$40×332
$\left(16\right)^y$(16)y $=$= $8\times\sqrt[3]{32}$8×332

 

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 $\sqrt[n]{x}=x^{\frac{1}{n}}$nx=x1n.

$\left(16\right)^y$(16)y $=$= $8\times\sqrt[3]{32}$8×332

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=1322.

 

Outcomes

ACMMM101

use exponential functions and their derivatives to solve practical problem

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