Year 10 - Exponential equations - Free practice

Year 10 Exponential Equations Worksheets

Same-base solving, index manipulation, growth/decay setup.

Quick answer

How do you solve exponential equations?

To solve a simple exponential equation, rewrite both sides as powers of the same base, then equate the exponents. For example, to solve 2^x = 32, write 32 as 2^5, giving x = 5. For harder cases, a substitution can reduce the equation to a quadratic.

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Six questions across Foundation, Standard and Extension, with full worked solutions, written to the Australian Curriculum.

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Worked examples

Year 10 exponential equations: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

  1. Example 1 · Solving by matching bases

    Solve 2x+1=322^{x+1} = 32.

    Show worked solution

    Key idea: Write both sides as powers of the same base, then equate the exponents.

    1. Write 32 as a power of 2: 32=2532 = 2^5, so the equation becomes 2x+1=252^{x+1} = 2^5
    2. Equate the exponents: x+1=5x + 1 = 5
    3. Solve for xx: x=4x = 4
    4. Check: 24+1=25=322^{4+1} = 2^5 = 32 ✓

    Answer: x=4x = 4

  2. Example 2 · Equation with a negative exponent

    Solve 32x−1=1273^{2x-1} = \dfrac{1}{27}.

    Show worked solution

    Key idea: Write the fraction as a power of 3 using a negative exponent, then equate the exponents.

    1. Write 127\dfrac{1}{27} as a power of 3: 127=3−3\dfrac{1}{27} = 3^{-3}, so the equation becomes 32x−1=3−33^{2x-1} = 3^{-3}
    2. Equate the exponents: 2x−1=−32x - 1 = -3
    3. Solve: 2x=−2  ⟹  x=−12x = -2 \implies x = -1
    4. Check: 32(−1)−1=3−3=1273^{2(-1)-1} = 3^{-3} = \dfrac{1}{27} ✓

    Answer: x=−1x = -1

  3. Example 3 · Exponential decay word problem

    A scientist stores a sample of a radioactive substance. Its mass in grams after tt years is given by M=800×2−t/5M = 800 \times 2^{-t/5}. After how many years will the mass first reach 100 grams?

    Show worked solution

    Key idea: Substitute the target mass into the equation, then write both sides as powers of 2 and equate the exponents.

    1. Substitute M=100M = 100: 100=800×2−t/5100 = 800 \times 2^{-t/5}
    2. Divide both sides by 800: 2−t/5=100800=182^{-t/5} = \frac{100}{800} = \frac{1}{8}
    3. Write 18\dfrac{1}{8} as a power of 2: 18=2−3\dfrac{1}{8} = 2^{-3}, so 2−t/5=2−32^{-t/5} = 2^{-3}
    4. Equate the exponents: −t5=−3-\frac{t}{5} = -3
    5. Solve: t=15t = 15
    6. Check: 800×2−15/5=800×2−3=800×18=100800 \times 2^{-15/5} = 800 \times 2^{-3} = 800 \times \dfrac{1}{8} = 100 ✓

    Answer: The mass reaches 100 grams after 15 years.

Common mistakes

Where students usually lose marks

Adding bases instead of equating exponents

2x=25  ⟹  x=5+2=72^x = 2^5 \implies x = 5 + 2 = 7

When both sides are written as the same base, only the exponents are set equal. The base is not involved in the equation you solve. Write 2x=25  ⟹  x=52^x = 2^5 \implies x = 5.

Misreading a fraction as a positive power

18=23\dfrac{1}{8} = 2^3 instead of 2−32^{-3}

A fraction 1an\dfrac{1}{a^n} equals a−na^{-n}, not ana^n. Always rewrite the fraction using a negative exponent before equating: 18=2−3\dfrac{1}{8} = 2^{-3}.

Forgetting to divide before equating

100=800×2−t/5  ⟹  2−t/5=100  ⟹  write 100 as a power of 2100 = 800 \times 2^{-t/5} \implies 2^{-t/5} = 100 \implies \text{write } 100 \text{ as a power of } 2

Isolate the exponential term first by dividing both sides by any coefficient. Here, divide by 800 to get 2−t/5=182^{-t/5} = \dfrac{1}{8} before converting to a matching base.

Equating exponents when bases are different

2x=32  ⟹  x=22^x = 3^2 \implies x = 2

You can only equate exponents after both sides are expressed as the same base. If the bases cannot be matched, this method does not apply and logarithms (studied later) are needed instead.

Try one yourself

Your turn

Solve 32x−1=813^{2x-1} = 81. What is the value of xx?

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About this worksheet

Why we built it

Year 10 exponential equations focus on the same-base technique - getting both sides expressed as a power of the same number and equating exponents. These worksheets give your student varied same-base problems plus the lead-up to logarithms in Year 10A.

What's covered

Sub-skills your student will practise

  • ✓Solving by writing both sides as the same base
  • ✓Working with negative and fractional exponents
  • ✓Setting up growth and decay equations
  • ✓Equations leading to a quadratic in disguise
  • ✓Word problems involving exponential change

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Frequently asked

Questions parents ask about exponential equations

What's an exponential equation?

An equation where the unknown is in the exponent - 2^x = 32, 3^(x + 1) = 81. Year 10 introduces the same-base method to solve these without logarithms.

What's the same-base method?

Rewrite both sides as a power of the same base, then equate exponents. For 2^x = 32: rewrite 32 as 2^5, so 2^x = 2^5, therefore x = 5. Works whenever both sides can be expressed in a common base.

What does Year 10 do when same-base doesn't work?

Mostly the questions are chosen so same-base does work. For cases where it doesn't (2^x = 7), logarithms are needed - that's Year 10A territory.

Where do exponential equations show up outside school?

Compound interest, population growth, radioactive decay, viral spread, cooling. Anywhere a quantity changes by a fixed proportion per unit of time, there's an exponential equation underneath.

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