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Indices

Indices

Solving equations: the unknown in the power

Equation Solving Paper 1 + 2 7 min read

On this page
  1. In this note
  2. Formula bank
  3. Make the bases the same
  4. Harder types: quadratics and simultaneous equations
  5. Ways they test it
  6. Mistakes to avoid
  7. Summary
  8. Practice
  9. Can I do this?
  10. Quiz

In this note

Syllabus 2.4: index equations - the unknown is IN the power
  • 2.4Solve equations like \(2^x=32\), \(3^{x-2}=81\), \(4^x=\tfrac1{64}\), \(32^x=2\) and \(5^{x+1}=25^x\).
  • 2.4Rewrite numbers, fractions, decimals and roots as a power of the same base.
  • 2.4Combine powers on one side first (\(3^x\times9^{x+1}\)), then equate the indices; harder ones lead to a quadratic or simultaneous equations.

Before you start: the rules of indices and negative and fractional indices. No logarithms are needed.

Formula bank

Equal powers of one baseLearn it\(a^{p}=a^{q}\ \Rightarrow\ p=q\)only when the bases are the SAME.Use it when the unknown is in the index.
Useful rewritesLearn it\(\tfrac1{a^n}=a^{-n}\), \(\sqrt a=a^{\frac12}\), \(1=a^{0}\)Use it when fractions, roots or 1 on one side.
Changing the baseLearn it\(4^{x+1}=(2^{2})^{x+1}=2^{2(x+1)}\)multiply the WHOLE index - keep the bracket.Use it when 4, 8, 9, 27, 25 … in an equation.

Make the bases the same

If two powers of the same base are equal, their indices are equal: \(2^x=2^5\) means \(x=5\). So every question is about getting BOTH sides into the form \(a^{\text{something}}\) with the same \(a\).

NumberPower of 2NumberPower of 3NumberPower of 5 / 10
4, 8, 16\(2^2,\ 2^3,\ 2^4\)9, 27, 81\(3^2,\ 3^3,\ 3^4\)25, 125\(5^2,\ 5^3\)
32, 64\(2^5,\ 2^6\)243, 729\(3^5,\ 3^6\)\(\tfrac1{25}=0.04\)\(5^{-2}\)
\(\tfrac18=0.125\)\(2^{-3}\)\(\tfrac1{27}\)\(3^{-3}\)\(0.001\)\(10^{-3}\)
\(\sqrt2\), \(\sqrt8\)\(2^{\frac12}\), \(2^{\frac32}\)\(\sqrt3\), \(\tfrac1{\sqrt3}\)\(3^{\frac12}\), \(3^{-\frac12}\)1\(5^0\)

General methodHow to solve any equation with the unknown in the power

  1. Pick the smallest prime base both sides are powers of (4, 8, 16 → 2; 9, 27, 81 → 3; 25, 125 → 5).
  2. Rewrite every number as a power of it: \(8=2^3\), \(\tfrac1{16}=2^{-4}\), \(\sqrt2=2^{\frac12}\), \(1=2^{0}\). Keep each index in a bracket: \(4^{x+1}=2^{2(x+1)}\).
  3. Collapse each side to ONE power with the rules (× → add indices, ÷ → subtract).
  4. Equate the indices and solve the ordinary equation that is left.
  5. Check by putting the answer back in.

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

Both are powers of 3.

  1. \(3^{2x}=3^{3(x-1)}\)
  2. \(2x=3x-3\)

Answer \(x=3\)

Example Solve \(3^x\times9^{x+1}=3^8\).

Collapse the left side first.

  1. \(3^x\times3^{2x+2}=3^{3x+2}\)
  2. \(3x+2=8\)

Answer \(x=2\)

\(4^x=8\) does NOT give \(x=2\): write \(2^{2x}=2^3\), so \(x=\tfrac32\).

Try it yourself

1
Solve \(4^x=\dfrac1{64}\).
Show answer

Answer −3

2
Solve \(3^{x-2}=81\).
Show answer

Answer 6

3
Solve \(32^x=2\).
Show answer

Answer \(\tfrac15\)

4
Solve \(5^{x+1}=25^x\).
Show answer

Answer 1

Harder types: quadratics and simultaneous equations

General methodHow to finish when the index equation is not linear

  1. Quadratic (an x² in an index): rearrange to = 0, factorise, give BOTH answers and check each.
  2. Two unknowns (two equations): turn each into an index equation, then solve the pair simultaneously.

Example Solve \(2^{x^2}=8^x\times16\).

Right side: \(2^{3x+4}\).

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

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

Example \(2^x\times4^y=128\) and \(\dfrac{3^x}{9^y}=\tfrac13\). Find \(x\) and \(y\).

One index equation from each.

  1. \(x+2y=7\)
  2. \(x-2y=-1\)
  3. add: \(2x=6\)

Answer \(x=3,\ y=2\)

Ways they test it

Type 1 Straight solve

"Find x: \(2^x=64\); \(5^x=0.04\)." (1-2 marks)

Spot it a number on the right.

  1. Write it as a power of the base.

For example \(5^x=0.04\)

  1. \(0.04=\tfrac1{25}=5^{-2}\)

Answer \(x=-2\)

Type 2 Unknown both sides

"Solve \(4^{2x-1}=8^x\)." (3 marks)

Spot it x in both indices.

  1. Base 2 both sides (M1).
  2. Linear equation (M1).

For example \(4^{2x-1}=8^x\)

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

Answer \(x=2\)

Type 3 Single power

"Write \(243\times27^n\) as a power of 3."

Spot it "as a single power".

  1. Add the indices.

Type 4 Point on a curve

"(5, 1024) lies on \(y=c^x\). Find c."

Spot it an exponential graph.

  1. \(c^5=1024\Rightarrow c=4\).

Type 5 Quadratic in the index

"Solve \(\dfrac{3^{x^2}}{27}=9^{x}\)." (4 marks)

Spot it an \(x^2\) in a power.

  1. Quadratic; two answers.

Mistakes to avoid

  • ✗ \(4^x=8\Rightarrow x=2\)✓ \(x=\tfrac32\)Bases must match before comparing.
  • ✗ \(4^{x+1}=2^{2x+1}\)✓ \(2^{2x+2}\)Multiply the whole index.
  • ✗ \(3^x\times9^x=27^{2x}\)✓ \(3^{3x}\)Never multiply the bases.
  • ✗ \(5^{x}=1\Rightarrow x=1\)✓ \(x=0\)\(1=5^0\).
  • ✗ \(x^2=3x+4\) → only \(x=4\)✓ \(x=4\) or \(-1\)Check both roots.

Summary

  • Same base both sides → equate indices.
  • Choose the smallest prime base (2, 3, 5, 10).
  • Fractions → negative index; roots → fractional index; 1 → index 0.
  • Collapse each side to one power; keep the whole index in a bracket.

If you see …

  • 4, 8, 16, 32, 64base 2.
  • an \(x^2\) in a powerquadratic: two answers.

Practice

Practice: easy, medium, hard

Easy

1
Solve \(2^x=64\).
Show answer

Answer 6

2
Solve \(3^x=\tfrac19\).
Show answer

Answer −2

3
Solve \(7^{x-4}=1\).
Show working
  1. \(1=7^0\)

Answer 4

Medium

1
Solve \(5^x=0.04\).
Guide me

What is it asking? x

What do we know? \(0.04=\tfrac1{25}\)

Plan \(5^{-2}\)

Show working
  1. \(5^x=5^{-2}\)

Answer −2

2
Solve \(4^{2x-1}=8^x\).
Guide me

What is it asking? x

What do we know? base 2

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

Show working
  1. \(4x-2=3x\)

Answer 2

3
Solve \(3^{x+2}=\dfrac1{27}\).
Guide me

What is it asking? x

What do we know? \(\tfrac1{27}=3^{-3}\)

Plan x + 2 = −3

Show working
  1. x = −5

Answer −5

Hard

1
\(2^{2x}\times8=4^5\). Find \(x\).
Guide me

What is it asking? x

What do we know? all base 2

Plan \(2^{2x+3}=2^{10}\)

Show working
  1. 2x + 3 = 10

Answer \(\tfrac72\)

PartAnswer / workingMarksGuidance
\(2^{2x+3}=2^{10}\)M1
2x + 3 = 10M1
3.5A1
Total: 3 marks
2
Find \(x\) when \(3^{x+1}\times27=9^{2x}\).
Guide me

What is it asking? x

What do we know? base 3

Plan \(3^{x+4}=3^{4x}\)

Show working
  1. x + 4 = 4x
  2. 3x = 4

Answer \(\tfrac43\)

PartAnswer / workingMarksGuidance
\(3^{x+4}=3^{4x}\)M1
x + 4 = 4xM1
\(\tfrac43\)A1
Total: 3 marks
3
Solve \(\dfrac{3^{x^2}}{27}=9^{x}\). Give both values.
Guide me

What is it asking? two values

What do we know? base 3

Plan \(x^2-3=2x\)

Show working
  1. \(x^2-2x-3=0\)
  2. \((x-3)(x+1)=0\)

Answer 3, −1

PartAnswer / workingMarksGuidance
quadraticM2
factorisedM1
3 and −1A1
Total: 4 marks

Can I do this?

Anything unticked: revise that section first. All ticked: take the quiz.

Quiz

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