Indices — Solving Equations

Section 1 of 3

Same Base

The key idea

To solve an index equation, write both sides as powers of the same base. Once the bases match, the powers must be equal — so you can solve for $x$.

Worked example — straight powers

Solve (i) $3^x = 9$; (ii) $2^x = 8$; (iii) $3^x = 243$.

(i) $3^x = 3^2 \Rightarrow x = 2$

(ii) $2^x = 2^3 \Rightarrow x = 3$

(iii) $3^x = 243 = 3^5 \Rightarrow x = 5$

$x = 2,\ 3,\ 5$

Section 2 of 3

With Roots

Turn roots into powers

Use $\sqrt{a} = a^{\frac{1}{2}}$ and the division rule to write the right-hand side as a single power.

Worked example — a root on the bottom

Solve $2^x = \dfrac{4}{\sqrt{2}}$.

$2^x = \dfrac{2^2}{2^{\frac{1}{2}}} = 2^{2 - \frac{1}{2}}$

$2^x = 2^{1\frac{1}{2}} \Rightarrow x = 1\tfrac{1}{2}$

$x = 1\tfrac{1}{2}$

Worked example — a root on top

Solve $6^x = 36\sqrt{6}$.

$6^x = 6^2 \cdot 6^{\frac{1}{2}} = 6^{2\frac{1}{2}}$

$x = 2\tfrac{1}{2}$

Section 3 of 3

With Brackets

Rewrite the base first

If the base isn’t prime (like $9$, $4$, $27$), rewrite it as a power ($9 = 3^2$), then use $(a^p)^q = a^{pq}$.

Worked example — a bracket equation

Solve $9^{x+1} = 27$.

$(3^2)^{x+1} = 3^3 \Rightarrow 3^{2x+2} = 3^3$

$2x + 2 = 3 \Rightarrow 2x = 1$

$x = \tfrac{1}{2}$

You try

Solve $5^x = 125$.

$125 = 5^3$.

$5^x = 5^3 \Rightarrow x = 3$

$x = 3$

You try

Solve $7^x = \dfrac{49}{\sqrt{7}}$.

$\dfrac{7^2}{7^{1/2}} = 7^{2 - \frac{1}{2}}$.

$7^x = 7^{1\frac{1}{2}} \Rightarrow x = 1\tfrac{1}{2}$

$x = 1\tfrac{1}{2}$

You try

Solve $3^x = \dfrac{\sqrt{3}}{9}$.

$\dfrac{3^{1/2}}{3^2} = 3^{\frac{1}{2} - 2}$.

$3^x = 3^{-1\frac{1}{2}} \Rightarrow x = -1\tfrac{1}{2}$

$x = -1\tfrac{1}{2}$

You try

Solve $4^{x-1} = 8$.

$4 = 2^2$ and $8 = 2^3$.

$(2^2)^{x-1} = 2^3 \Rightarrow 2^{2x-2} = 2^3$

$2x - 2 = 3 \Rightarrow 2x = 5 \Rightarrow x = \tfrac{5}{2}$

$x = \tfrac{5}{2}$

That’s Indices — and that’s Algebra complete!

Same base, roots as fractional powers, and rewriting brackets. You’ve now got the whole Algebra topic including indices.