Indices & standard form · Lesson 2 of 3

Equations with indices

Solving equations where the unknown is in the index: make the bases the same and compare indices, factor out a common power, spot a hidden quadratic, undo a power on the unknown, and solve two such equations together.

20 minYou should already know: Number foundations & fractions
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In an exponential equation the unknown is in the index, as in 2x=322^x = 32. The main method is one idea:

If ap=aqa^p = a^q (with the same base aa, not 0 or 1), then p=qp = q.

So write both sides as powers of the same base, then set the indices equal. 2x=32=252^x = 32 = 2^5, so x=5x = 5.

More: make the bases the same

Worked example · NECO 2023

NECO 2023 · Paper 2 · Q4 (a)

Solve the equation 272x−1×(13)−(3x+2)=9x+327^{2x - 1} \times \left(\frac13\right)^{-(3x + 2)} = 9^{x + 3}.

  1. Everything as a power of 3

    27=3327 = 3^3, 13=3−1\frac13 = 3^{-1} and 9=329 = 3^2.

    Think first. Write 27, 13\frac13 and 9 as powers of 3.

  2. Multiply out the indices

    272x−1=36x−327^{2x - 1} = 3^{6x - 3}, (13)−(3x+2)=33x+2\left(\frac13\right)^{-(3x + 2)} = 3^{3x + 2} and 9x+3=32x+69^{x + 3} = 3^{2x + 6}.

  3. Add the indices on the left

    3(6x−3)+(3x+2)=39x−13^{(6x - 3) + (3x + 2)} = 3^{9x - 1}, so 39x−1=32x+63^{9x - 1} = 3^{2x + 6}.

    Think first. Multiplying powers of 3: what happens to the indices?

  4. Compare indices

    9x−1=2x+69x - 1 = 2x + 6, so 7x=77x = 7 and x=1x = 1.

When the terms are added: factor out the power

3x+3x+13^x + 3^{x + 1} can’t be combined into one power directly, but 3x+1=3×3x3^{x + 1} = 3 \times 3^x. Take 3x3^x out as a common factor.

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Equations that hide a quadratic

4x=(22)x=(2x)24^x = (2^2)^x = (2^x)^2, so an equation with 4x4^x and 2x2^x in it is a quadratic in disguise. Write y=2xy = 2^x, solve for yy, then turn each yy back into xx. For 4x−6(2x)+8=04^x - 6(2^x) + 8 = 0:

4ˣ = (2²)ˣ = (2ˣ)²(2ˣ)² − 6(2ˣ) + 8 = 0y² − 6y + 8 = 0
A quadratic in disguiseLet y = 2ˣ

y2−6y+8=0y^2 - 6y + 8 = 0 gives (y−2)(y−4)=0(y - 2)(y - 4) = 0, so 2x=22^x = 2 or 2x=42^x = 4, and x=1x = 1 or x=2x = 2. A negative value of yy gives no answer, because 2x2^x is always positive.

More: hidden quadratics

When the unknown is the base

In x23=9x^{\frac23} = 9 the unknown is the base, not the index. Undo the power by raising both sides to its reciprocal: x=932=27x = 9^{\frac32} = 27. A root works the same way: from 2+x+13=52 + \sqrt[3]{x + 1} = 5, get the root alone (x+13=3\sqrt[3]{x + 1} = 3), then cube both sides (x+1=27x + 1 = 27, so x=26x = 26).

More: the unknown in the base

Two equations together

When two exponential equations share two unknowns, turn each into a simple equation in the indices, then solve the pair as simultaneous equations.

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q6 (a)

Given that 2m×(18)n=1282^m \times \left(\frac18\right)^n = 128 and 4m÷2−4n=1164^m \div 2^{-4n} = \frac{1}{16}, find the value of (m−n)(m - n).

  1. The first equation in base 2

    2m×2−3n=272^m \times 2^{-3n} = 2^7, so m−3n=7m - 3n = 7.

    Think first. 18=2?\frac18 = 2^{?} and 128=2?128 = 2^{?}

  2. The second equation in base 2

    4m=22m4^m = 2^{2m}, and dividing by 2−4n2^{-4n} adds 4n4n: 22m+4n=2−42^{2m + 4n} = 2^{-4}, so 2m+4n=−42m + 4n = -4, or m+2n=−2m + 2n = -2.

  3. Solve the pair

    Subtract: (m−3n)−(m+2n)=7−(−2)(m - 3n) - (m + 2n) = 7 - (-2), so −5n=9-5n = 9 and n=−95n = -\frac95. Then m=7+3n=85m = 7 + 3n = \frac85.

  4. Answer the question asked

    m−n=85+95=175=325m - n = \frac85 + \frac95 = \frac{17}{5} = 3\frac25.

More: two equations together

Your turn

WAEC 2021 · Paper 1 · Q10

If 16×2(x+1)=4x×8(1−x)16 \times 2^{(x + 1)} = 4^x \times 8^{(1 - x)}, find the value of xx.

Worked solution (try it first)
  1. Write everything as a power of 2.
  2. Left: 16×2x+1=24×2x+116 \times 2^{x + 1} = 2^4 \times 2^{x + 1}
    =2x+5= 2^{x + 5}.
  3. Right: 4x=22x4^x = 2^{2x} and 81−x=23−3x8^{1 - x} = 2^{3 - 3x}, so the right side is 22x+3−3x=23−x2^{2x + 3 - 3x} = 2^{3 - x}.
  4. Equate the powers: x+5=3−xx + 5 = 3 - x, so 2x=−22x = -2 and x=−1x = -1, option D.

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