About this video
More advanced exponential equations, including using factorising and substitution when the bases can't be matched directly.
What it covers
- Same bases: equate the exponents
- Decimals and fractions as powers (0,25 = 2⁻²)
- Taking out a common factor such as 2ˣ
- The k-method: let 3ˣ = k, factorise, then resubstitute
- When there is no solution: 3ˣ can never be negative
Extra worked examples
✍️ Extra example (common factor): solve 3x + 2 + 3x = 90
- 13x · 32 + 3x = 90Split the exponent so the common factor shows.
- 23x(9 + 1) = 90 → 3x · 10 = 90Take out the common factor 3ˣ.
- 33x = 9 = 32 → x = 2Divide by 10, make the bases the same, equate the exponents.
✍️ Extra example (the k-method): solve 22x − 6 · 2x + 8 = 0
- 1Let 2x = k, so 22x = k².Any letter except x — the substitution makes it look like a trinomial.
- 2k² − 6k + 8 = 0 → (k − 2)(k − 4) = 0Factorise the trinomial.
- 32x = 2 → x = 1, or 2x = 4 = 22 → x = 2Resubstitute — you are solving for x, not k.
Exercises to try
Work each one out, then click to reveal the answer.
- 1Solve 9x + 1 = 27.
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32x + 2 = 33 → 2x + 2 = 3 → x = ½ - 2Solve 2x + 3 − 2x = 56.
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2x(8 − 1) = 56 → 2x = 8 = 23 → x = 3 - 3Solve 52x − 6 · 5x + 5 = 0.
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(k − 5)(k − 1) = 0 with k = 5x → 5x = 5 → x = 1, or 5x = 1 = 50 → x = 0 - 4Solve 2x − 3 = 4/2x.
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× 2x: k² − 3k − 4 = 0 → (k − 4)(k + 1) = 0 → 2x = 4 → x = 2; 2x = −1 has no solution (a power of 2 is never negative) - 5Solve (0,5)x = 16.
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0,5 = 1/2 = 2−1, so 2−x = 24 → −x = 4 → x = −4
