How do you solve an equation like 4^x minus 5 times 2^x plus 4 equals 0?

Let u equal 2 to the power x. Since 4 to the power x equals u squared, the equation becomes u squared minus 5u plus 4 equals 0, which factorises to give u equals 1 or u equals 4. Then convert back: 2 to the power x equals 1 gives x equals 0, and 2 to the power x equals 4 gives x equals 2.

How do you solve an equation like 4^x minus 5 times 2^x plus 4 equals 0?

Let u equal 2 to the power x. Since 4 to the power x equals u squared, the equation becomes u squared minus 5u plus 4 equals 0, which factorises to give u equals 1 or u equals 4. Then convert back: 2 to the power x equals 1 gives x equals 0, and 2 to the power x equals 4 gives x equals 2.

Year 12 Methods (Unit 3 & 4) Exponential and logarithmic functions

Revision of exponential equations

Q: When do you need logarithms to solve an exponential equation?

You need logarithms when the number on the other side is not a whole power of the base, so equal bases are impossible. For 2 to the power x equals 10, take logs of both sides to get x equals log 10 divided by log 2, which is about 3.32.

Q: What is the same-base method?

If two powers of the same positive base a (not equal to 1) are equal, then their indices are equal: a to the power m equals a to the power n means m equals n. So you rewrite every term as a power of one common base, then set the exponents equal and solve the resulting equation.

Q: Can you always write both sides with the same base?

No. It works when both numbers are powers of a common base, such as 8, 16 and 2 all being powers of 2. When one side is not a whole power of the base, for example 2 to the power x equals 20, you must solve with logarithms or with technology instead.

Q: Why do you reject a negative value of u = a^x?

Because a to the power x is always positive for a positive base a, the substitution u equals a to the power x can never be zero or negative. So when a quadratic in u gives a negative root, that root has no matching value of x and is discarded.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), an exponential equation has the unknown in the exponent. The main method is to write both sides as a power of the same base and equate the indices — extended here to equations that need several bases converted to a common base, equations that reduce to a quadratic in \(a^x\), and equations solved with logarithms when the bases cannot be matched, ready for the calculus of Unit 3.

An exponential equation is an equation in which the unknown appears in an exponent, such as \(2^{x}=32\) or \(4^{x}-5\times 2^{x}+4=0\). Solving it means finding the value(s) of \(x\) that make it true.

The key tool is the same-base principle: for a positive base \(a\) (\(a\neq 1\)),

\(a^{m}=a^{n}\ \Longleftrightarrow\ m=n.\)

So you rewrite every term as a power of one common base using the index laws, then simply equate the indices and solve. When the number on the other side is not a whole power of the base (for example \(2^{x}=10\)), equal bases are impossible, and you take logarithms of both sides instead. When a power such as \(a^{x}\) appears squared, the substitution \(u=a^{x}\) turns the problem into a quadratic.

Key idea. Same base \(\Rightarrow\) equate indices: \(a^{m}=a^{n}\Rightarrow m=n\) (\(a>0,\ a\neq 1\)). No common base possible \(\Rightarrow\) take logs. A power squared \(\Rightarrow\) let \(u=a^{x}\).

Same base: \(2^{x}=8=2^{3}\), so \(x=3\) — where \(y=2^{x}\) meets \(y=8\)

No common base: \(2^{x}=10\) gives \(x=\log_{2}10\approx 3.32\) — solve with logs

The same-base principle (\(a>0,\ a\neq 1\)):

\[a^{m}=a^{n}\ \Longleftrightarrow\ m=n\]

a^{m} = a^{n} \Leftrightarrow m = n

The index laws used to reach a common base:

\[a^{m}\times a^{n}=a^{m+n},\quad \dfrac{a^{m}}{a^{n}}=a^{m-n},\quad (a^{m})^{n}=a^{mn},\quad a^{-n}=\dfrac{1}{a^{n}},\quad a^{0}=1\]

a^{m} \times a^{n} = a^{m + n}

When the bases cannot be matched, take logarithms:

\[a^{x}=k \ \Rightarrow\ x=\dfrac{\log k}{\log a}=\log_{a}k\]

x = \frac{\log k}{\log a}

Quadratic form. If \(a^{x}\) appears squared, let \(u=a^{x}\); then \(a^{2x}=(a^{x})^{2}=u^{2}\), and the equation becomes a quadratic in \(u\). Solve for \(u\), discard any \(u\le 0\) (since \(a^{x}>0\)), then convert each valid \(u\) back to \(x\).

How to solve an exponential equation

Try a common base. Rewrite every term as a power of one base using the index laws (e.g. \(8=2^{3}\), \(9=3^{2}\)).
Equate the indices. Once each side is a single power of that base, set the exponents equal and solve the resulting equation.
Spot a quadratic. If a power such as \(2^{x}\) appears squared (as \(4^{x}\)), let \(u=2^{x}\), solve the quadratic, discard \(u\le 0\), then convert back.
Otherwise use logs. If no common base works, take logarithms of both sides: \(a^{x}=k\Rightarrow x=\dfrac{\log k}{\log a}\).

Modelling. Growth and decay problems lead to exponential equations. To find the time in \(N=N_{0}\,r^{t}\), substitute the target value and solve for \(t\) — by a common base if \(r^{t}\) is a neat power, otherwise with logarithms.

Example 1 — Same base

Solve \(2^{x+1}=8\).

Solution

Write \(8\) as a power of \(2\), then equate the indices.

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

2^{x + 1} = 2^{3}, x = 2

Example 2 — Several bases

Solve \(4^{x+2}\times 8^{1-x}=2\).

Solution

Write \(4\), \(8\) and \(2\) as powers of \(2\).

\(2^{2(x+2)}\times 2^{3(1-x)}\)	\(=\)	\(2^{1}\)
\(2^{\,7-x}\)	\(=\)	\(2^{1}\)
\(7-x\)	\(=\)	\(1\)
\(x\)	\(=\)	\(6\)

2^{7 - x} = 2^{1}, x = 6

Example 3 — Quadratic in \(2^{x}\)

Solve \(4^{x}-5\times 2^{x}+4=0\) by letting \(u=2^{x}\).

Solution

Since \(4^{x}=(2^{x})^{2}=u^{2}\):

\(u^{2}-5u+4\)	\(=\)	\(0\)
\((u-1)(u-4)\)	\(=\)	\(0\)
\(u\)	\(=\)	\(1\) or \(4\)

Convert back: \(2^{x}=1\Rightarrow x=0\); \(2^{x}=4\Rightarrow x=2\).

x = 0 or x = 2

Example 4 — Solve with logs

Solve \(2^{x}=10\), correct to \(2\) decimal places.

Solution

\(10\) is not a power of \(2\), so take logarithms of both sides.

\(\log\!\left(2^{x}\right)\)	\(=\)	\(\log 10\)
\(x\log 2\)	\(=\)	\(\log 10\)
\(x\)	\(=\)	\(\dfrac{\log 10}{\log 2}\approx 3.32\)

x = \frac{\log 10}{\log 2} \approx 3.32

Common pitfalls

You can only equate indices when the bases match. Rewrite every term as a power of the same base first; you cannot jump from \(2^{x}=3^{5}\) to \(x=5\), because the bases differ.

Convert \(u\) back to \(x\). After letting \(u=2^{x}\) and solving the quadratic, the values of \(u\) are not the answers — solve \(2^{x}=u\) for each valid \(u\) to find \(x\).

Reject non-positive \(u\). Because \(a^{x}>0\) always, a root such as \(u=-5\) has no solution; discard it. Missing this either invents a wrong answer or discards a correct one.

Frequently asked questions

How do you solve an exponential equation?

Where possible, write both sides as a power of the same base and equate the indices — e.g. \(2^{x}=32=2^{5}\) gives \(x=5\). If the sides cannot share a base, take logarithms instead.

When do you need logarithms to solve an exponential equation?

When the other side is not a whole power of the base, so equal bases are impossible. For \(2^{x}=10\), \(x=\dfrac{\log 10}{\log 2}\approx 3.32\).

How do you solve an equation like 4^x - 5(2^x) + 4 = 0?

Let \(u=2^{x}\). Then \(4^{x}=u^{2}\), so \(u^{2}-5u+4=0\), giving \(u=1\) or \(4\). Convert back: \(x=0\) or \(x=2\).

What is the same-base method?

If \(a^{m}=a^{n}\) for a positive base \(a\neq 1\), then \(m=n\). Rewrite each term as a power of one common base, then set the exponents equal and solve.

Can you always write both sides with the same base?

No. It only works when both numbers are powers of a common base (like \(8,16,2\) with base \(2\)). Otherwise solve with logarithms or technology.

Why do you reject a negative value of u = a^x?

Because \(a^{x}\) is always positive, so \(u=a^{x}\) can never be zero or negative; a negative root of the quadratic has no matching \(x\).

Practice 20 questions

Practice questions

Every question with a fully worked solution.

Start practising

← Previous subtopic

The exponential functionf(x)=ex.

Next subtopic →

Logarithms and the logarithm laws

More in Exponential and logarithmic functions