To solve an exponential equation by a common base, rewrite both sides as powers of the same number, then set the powers equal. This turns an equation with x in the exponent into an ordinary linear (or quadratic) equation.
It is one of the skills in algebraic equations and inequalities, and the full logarithm approach appears in exponential and logarithmic reasoning.
What are the index laws doing here?
Two laws carry almost all of the work. (a^m)^n = a^(mn), which lets you rewrite 4^x as (2²)^x = 2^(2x). And a^m × a^n = a^(m+n), which lets you combine products such as 2^x × 2^3 into 2^(x+3).
The key fact is that if the base is the same on both sides, the powers must match. So 2^(2x) = 2^(3x − 3) means 2x = 3x − 3.
Method, step by step
- Pick a common base, usually the smallest base that both sides are powers of (2, 3, 5 and so on).
- Rewrite each side as a single power of that base, using index laws.
- Equate the powers and drop the base.
- Solve the resulting equation.
- Check by substituting back into the original equation.
Worked example
Solve 4^x = 8^(x − 1).
Step 1, choose base 2: 4 = 2² and 8 = 2³.
Step 2, rewrite: (2²)^x = (2³)^(x − 1), so 2^(2x) = 2^(3(x − 1)) = 2^(3x − 3).
Step 3, equate the powers: 2x = 3x − 3.
Step 4, solve: −x = −3, so x = 3.
Check: 4³ = 64 and 8² = 64. Both sides agree.
Answer: x = 3.
The mistake to watch for
The common slip is to treat the different bases as if they could be compared directly.
Mistaken working: 4^x = 8^(x − 1), so 4x = 8(x − 1)
The student multiplied each exponent by its own base. That gives x = 2, and the check fails: 4² = 16 but 8^1 = 8.
An exponent and a base are different things. The base must be changed first, through index laws, so that both sides share it. Only then can you compare the powers.
The substitution check catches this error straight away.
Check yourself
Try these, then open each answer.
1. Solve 3^(2x − 1) = 27.
Show answer
27 = 3³, so 2x − 1 = 3, giving x = 2. Check: 3^(4 − 1) = 3³ = 27.
x = 2
2. Solve 9^x = 27^(x − 2).
Show answer
Base 3: (3²)^x = (3³)^(x − 2), so 2x = 3(x − 2) = 3x − 6, giving x = 6.
Check: 9^6 = 3^12 and 27^4 = 3^12.
x = 6
3. Solve 2^(x + 3) = 1/16.
Show answer
1/16 = 2^(−4), so x + 3 = −4, giving x = −7. Check: 2^(−4) = 1/16.
x = −7
Where this leads next
If a product of terms appears instead of a single power, the same rewriting pattern applies after factorising. Continue with stating restrictions before manipulating fractions, then try the mixed practice set. The non-calculator working trainer is useful for practising exact power checks without a calculator.
If index laws feel solid in isolation but fall apart inside longer questions, that is something our teachers can trace step by step in online one-to-one Additional Mathematics tuition.