Equations with indices

In an exponential equation the unknown is in the index, as in $2^x = 32$. The main method is one idea:

If $a^p = a^q$ (with the same base $a$, not 0 or 1), then $p = q$.

So write both sides as powers of the same base, then set the indices equal. For $2^x = 32$: $32 = 2^5$, so $2^x = 2^5$ and $x = 5$.

When the terms are added: factor out the power

$3^x + 3^{x + 1}$ can’t be combined into one power directly. But $3^{x + 1} = 3 \times 3^x$, so you can take $3^x$ out as a common factor.

Equations that hide a quadratic

$4^x = (2^2)^x = (2^x)^2$. So an equation with both $4^x$ and $2^x$ in it is a quadratic↺ in disguise. Write $y = 2^x$, solve for $y$, then turn each $y$ back into $x$. For $4^x - 6(2^x) + 8 = 0$:

• $y^2 - 6y + 8 = 0$ factorises to $(y - 2)(y - 4) = 0$, so $y = 2$ or $y = 4$.
• $2^x = 2$ gives $x = 1$, and $2^x = 4$ gives $x = 2$.

If you get a negative value of $y$, it gives no answer, because $2^x$ is always positive.

When the unknown is the base

In $x^{\frac23} = 9$ the unknown is the base, not the index. Undo the power by raising both sides to its reciprocal, the fraction turned upside down:

• raise both sides to the power $\frac32$: ${x = 9^{\frac32}}$;
• root first, then power: ${\left(\sqrt9\right)^3 = 3^3 = 27}$.

A root works the same way. From $2 + \sqrt[3]{x + 1} = 5$, get the root alone: $\sqrt[3]{x + 1} = 3$. Then cube both sides: $x + 1 = 27$, so $x = 26$.

Two equations together

When two exponential equations share two unknowns, turn each into a simple equation in the indices, then solve the pair as simultaneous equations↺.

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