How do you solve three to the x equals one over eighty one?

Write one over eighty one as three to the minus four. Then three to the x equals three to the minus four, so x equals minus four.

How do you solve three to the x equals one over eighty one?

Write one over eighty one as three to the minus four. Then three to the x equals three to the minus four, so x equals minus four.

Year 11 Maths - Methods (Unit 1 and Unit 2) Exponential and logarithmic functions

Solving Exponential Equations and Inequalities

Q: How do you solve an exponential equation without logarithms?

Write both sides as powers of the same base, then set the indices equal. For two to the x equals thirty two, which is two to the fifth, the indices give x equals five.

Q: How do you get a common base?

Rewrite each side using the index laws. For example twenty seven is three cubed, one ninth is three to the minus two, and a to the m all to the x is a to the m times x. Keep going until both sides share one base.

Q: What happens to the sign in an exponential inequality?

For a base greater than one the function is increasing, so comparing the indices keeps the inequality the same way round. Two to the two x minus one is greater than or equal to two to the five gives two x minus one is greater than or equal to five.

Q: Why can an exponential equation have two solutions?

If equating the indices produces a quadratic, such as x squared minus three x equals four, it factorises to give two values. Here x equals four or x equals minus one.

20 practice questions 0 video lessons Theory + worked examples

Learn to solve exponential equations and inequalities for Victorian Year 11 Mathematical Methods (VCAA). When the unknown is in the power, writing both sides to a common base lets you equate the indices and solve.

You will learn to rewrite equations to a common base, handle negative and reciprocal targets, solve inequalities where the base is greater than one, and manage equations that become a quadratic — all without logarithms.

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Practice questions

Every question with a fully worked solution.

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Theory

In Year 11 Mathematical Methods (Unit 2), you solve an exponential equation such as \(2^{x}=32\) by writing both sides to a common base and then equating the indices — no logarithms needed. This page covers common-base equations, negative and reciprocal targets, inequalities, and equations that reduce to a quadratic.

An exponential equation has the unknown in the index, such as \(2^{x}=32\). If both sides can be written with the same base, then the powers are equal exactly when their indices are equal: \(a^{x}=a^{k}\Rightarrow x=k\).

The method is to rewrite each side to a common base using the index laws, then equate the indices and solve the resulting equation. For an inequality with base \(r>1\), the same step keeps the direction the same, because \(r^{x}\) is increasing. As a special case \(a^{x}=1\) gives \(x=0\).

Same base, then compare indices. Once both sides read \(a^{\square}=a^{\triangle}\), just set \(\square=\triangle\) and solve.

Reading a solution: \(y=2^{x}\) meets \(y=8\) at \(x=3\), so \(2^{x}=8\) gives \(x=3\).

An inequality solution shown on a number line: \(x\ge 3\) (filled dot, ray to the right).

The key step — equal bases mean equal indices:

\[a^{x}=a^{k}\quad\Longrightarrow\quad x=k\qquad(a>0,\ a\neq1)\]

a^{x} = a^{k} \Rightarrow x = k

Useful rewrites for finding a common base:

\[\dfrac{1}{a^{n}}=a^{-n},\qquad \sqrt[n]{a}=a^{1/n},\qquad (a^{m})^{x}=a^{mx}\]

\frac{1}{a^{n}} = a^{- n}

Inequality with base \(r>1\): \(r^{x}\) increases, so \(r^{f(x)}\ge r^{g(x)}\) becomes \(f(x)\ge g(x)\) — the inequality sign is unchanged.

How to solve \(a^{x}=b\) by equating indices

Common base: rewrite both sides as powers of the same base, using \(\dfrac{1}{a^{n}}=a^{-n}\) and \((a^{m})^{x}=a^{mx}\) as needed.
Equate the indices: once the bases match, set the indices equal (for an inequality with base \(r>1\), keep the same direction).
Solve: solve the resulting linear, quadratic or inequality statement for \(x\).

Example 1 — A common-base equation

Solve \(2^{x}=32\).

Solution

Write \(32\) as a power of \(2\):

\(32\)

\(=\)

\(2^{5}\)

So the equation reads powers of \(2\) — equate the indices:

\(2^{x}\)	\(=\)	\(2^{5}\)
\(x\)	\(=\)	\(5\)

\(x=5\).

x = 5

Example 2 — A negative, reciprocal target

Solve \(27^{x}=\dfrac{1}{9}\).

Solution

Common base \(3\): \(27=3^{3}\) and \(\dfrac{1}{9}=3^{-2}\):

\((3^{3})^{x}\)	\(=\)	\(3^{-2}\)
\(3^{3x}\)	\(=\)	\(3^{-2}\)

Equate the indices and solve:

\(3x\)	\(=\)	\(-2\)
\(x\)	\(=\)	\(-\dfrac{2}{3}\)

\(x=-\dfrac{2}{3}\).

x = - \frac{2}{3}

Example 3 — An exponential inequality

Solve \(2^{\,2x-1}\ge 32\).

Solution

Write \(32\) as a power of \(2\):

\(32\)

\(=\)

\(2^{5}\)

Base \(2>1\) is increasing, so keep the direction and compare indices:

\(2^{\,2x-1}\)	\(\ge\)	\(2^{5}\)
\(2x-1\)	\(\ge\)	\(5\)

Solve the linear inequality:

\(2x\)	\(\ge\)	\(6\)
\(x\)	\(\ge\)	\(3\)

\(x\ge 3\).

x \geq 3

Example 4 — Reducing to a quadratic

Solve \(2^{\,x^{2}-3x}=16\).

Solution

Write \(16\) as a power of \(2\), then equate indices:

\(16\)	\(=\)	\(2^{4}\)
\(x^{2}-3x\)	\(=\)	\(4\)

Rearrange to a quadratic and factorise:

\(x^{2}-3x-4\)	\(=\)	\(0\)
\((x-4)(x+1)\)	\(=\)	\(0\)

Read the two solutions:

\(x\)

\(=\)

\(4\quad\text{or}\quad x=-1\)

\(x=4\) or \(x=-1\).

x = 4 or x = - 1

Common pitfalls

Not matching the bases first. You can only equate indices once both sides share a base. \(2^{x}=32\) works because \(32=2^{5}\); compare indices only after that.

Mishandling a reciprocal. \(\dfrac{1}{9}=3^{-2}\), so the index is \(-2\); a reciprocal makes the index negative, it does not make it a fraction on its own.

Flipping an inequality by mistake. For base \(r>1\) the function is increasing, so the inequality direction is preserved when you compare indices.

Dropping a solution. When the indices form a quadratic, expect two answers; solve \(x^{2}-3x-4=0\) fully to get \(x=4\) and \(x=-1\).

Frequently asked questions

How do you solve an exponential equation without logarithms?

Write both sides as powers of the same base, then equate the indices. For \(2^{x}=32=2^{5}\), the indices give \(x=5\).

How do you get a common base?

Rewrite each side using the index laws, for example \(27=3^{3}\), \(\dfrac{1}{9}=3^{-2}\) and \((a^{m})^{x}=a^{mx}\), until both sides share one base.

What happens to the sign in an exponential inequality?

For a base \(r>1\) the function is increasing, so comparing indices keeps the inequality the same way round: \(2^{2x-1}\ge 2^{5}\) gives \(2x-1\ge 5\).

How do you solve 3^x = 1/81?

Write \(\dfrac{1}{81}=3^{-4}\), so \(3^{x}=3^{-4}\) and \(x=-4\).

Why can an exponential equation have two solutions?

If equating the indices produces a quadratic, such as \(x^{2}-3x=4\), it factorises to give two values, here \(x=4\) and \(x=-1\).

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