Equations

New Zealand

Level 6 - NCEA Level 1

Lesson

In this chapter, we are going to look at how to solve simple exponential equations, where the unknown value we are trying to find is in the power.

For example, let's say we wanted to solve the equation $2^{4x}=2^8$24`x`=28.

We can see that the base terms on both sides equal $2$2. Hence, we can say that $4x=8$4`x`=8 This is because that the only way $2^{\text{something }}$2something can equal $2^{\text{something else }}$2something else is if the powers are both the same. From here we can solve this as a regular algebraic equation, where we get the solution $x=2$`x`=2.

Sometimes we can express numbers in terms of a base number with an exponent. For example, $25$25 can be expressed as $5^2$52, $64$64 can be expressed as $4^3$43 and so on.

Let's look at an equation where we need to rewrite a number to solve an exponential equation. For example, let's solve the equation $2^{3x-3}=8$23`x`−3=8.

At first glance, it seems like it's difficult to solve because we don't have like bases. However, we can rewrite $8$8 as $2^3$23. Now we have like base terms!

$2^{3x-3}$23x−3 |
$=$= | $8$8 |

$2^{3x-3}$23x−3 |
$=$= | $2^3$23 |

$3x-3$3x−3 |
$=$= | $3$3 |

$3x$3x |
$=$= | $6$6 |

$x$x |
$=$= | $2$2 |

Let's look through some more examples so we master solving exponential equations!

Solve the equation $5^{-3x-1}=3125$5−3`x`−1=3125 for $x$`x`.

Solve the equation $\left(2^2\right)^y=2^3$(22)`y`=23 for $y$`y`.

Solve the equation $\left(2^2\right)^{x+7}=2^3$(22)`x`+7=23 for $x$`x`.

Form and solve linear equations and inequations, quadratic and simple exponential equations, and simultaneous equations with two unknowns

Apply algebraic procedures in solving problems