Year 10 and 10A Indices, exponentials and logarithms: free practice test with answers

Sample indices, exponentials and logarithms questions with answers

Try each one before you open the answer.

1. Review of index laws

   Simplify $3a^4 \times 5a^3$.

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   Key idea: When multiplying terms with the same base, add the exponents: $a^m \times a^n = a^{m+n}$.

   1. Multiply the coefficients: $3 \times 5 = 15$.
   2. Apply the product law to the powers of $a$: $a^4 \times a^3 = a^{4+3} = a^7$.
   3. Write the result: $15a^7$.

   Answer: $15a^7$

2. Negative indices

   Evaluate $4^{-2}$, giving your answer as a fraction or a decimal.

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   Key idea: $a^{-n} = \frac{1}{a^{n}}$.

   1. $4^{-2} = \frac{1}{4^{2}}$.
   2. $= \frac{1}{16} = 0.0625$.

   Answer: $\frac{1}{16}$

3. Fractional indices

   A storage container is a cube with a volume of $200$ cubic metres. The side length of a cube with volume $V$ is given by $s = V^{\frac{1}{3}}$. Write the exact side length of this container in surd form, and then evaluate it correct to two decimal places.

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   Key idea: $V^{\frac{1}{3}}$ means the cube root of $V$, so $V^{\frac{1}{3}} = \sqrt[3]{V}$.

   1. Substitute $V = 200$: $s = 200^{\frac{1}{3}} = \sqrt[3]{200}$
   2. Write $200 = 8 \times 25$, so: $\sqrt[3]{200} = \sqrt[3]{8 \times 25} = \sqrt[3]{8} \times \sqrt[3]{25} = 2\sqrt[3]{25}$
   3. Evaluate numerically: $2\sqrt[3]{25} \approx 2 \times 2.9240 \approx 5.85 \text{ m}$

   Answer: Exact side length is $2\sqrt[3]{25}$ m, which is approximately $5.85$ m.

4. Exponential equations

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

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

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

   Answer: $x = 4$

5. Exponential growth and decay

   A scientist measures the mass of a radioactive sample. At time $t = 0$ the mass is $80$ grams. The mass halves every hour, so the mass in grams after $t$ hours is modelled by $M = 80 \times \left(\dfrac{1}{2}\right)^t$.

   (a) Write the mass when $t = 0$, $t = 1$, $t = 2$ and $t = 3$. (b) State the horizontal asymptote of the graph of $M$ against $t$ and explain what it means in this context. (c) After how many hours is the mass first below $15$ grams?

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   Key idea: $M = 80 \times \left(\dfrac{1}{2}\right)^t$ is an exponential decay model. The base $\tfrac{1}{2} < 1$, so $M$ decreases and approaches $0$ but never reaches it.

   1. Evaluate at each time: $\begin{aligned} t=0:& M = 80 \times 1 = 80 \text{ g}\\ t=1:& M = 80 \times \tfrac{1}{2} = 40 \text{ g}\\ t=2:& M = 80 \times \tfrac{1}{4} = 20 \text{ g}\\ t=3:& M = 80 \times \tfrac{1}{8} = 10 \text{ g} \end{aligned}$
   2. The horizontal asymptote is $M = 0$. This means the mass gets closer and closer to zero grams but, according to the model, never actually reaches zero.
   3. From step 1, at $t = 2$ the mass is $20$ g (not yet below $15$ g), and at $t = 3$ the mass is $10$ g (below $15$ g). So the mass is first below $15$ grams after $3$ hours.

   Answer: (a) $80$ g, $40$ g, $20$ g, $10$ g. (b) Asymptote $M = 0$; the mass never reaches zero. (c) After $3$ hours.

6. Compound interest

   Calculate the value of an investment of $8000 after 3 years at an interest rate of 5% per annum, compounded annually.

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   Key idea: Use the compound interest formula $A = P(1 + r)^n$, where $P$ is the principal, $r$ is the annual interest rate as a decimal, and $n$ is the number of compounding periods.

   1. Identify the values: $P = 8000$, $r = 0.05$, $n = 3$.
   2. Substitute into the formula:\

   $$A = 8000(1 + 0.05)^3$$

   3. Simplify:\

   $$A = 8000 \times (1.05)^3 = 8000 \times 1.157625 = 9261.00$$

   4. Find the interest earned: $9261.00 - 8000 = 1261.00$.

   Answer: The investment is worth $9261.00, so the compound interest earned is $1261.00.

7. Logarithmic scales

   The pH scale is a base-10 logarithmic scale defined by $\text{pH} = -\log_{10}[\text{H}^+]$, where $[\text{H}^+]$ is the hydrogen ion concentration in moles per litre. A solution has a pH of 3. Find its hydrogen ion concentration.

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   Key idea: On a base-10 log scale, if the log scale reading is $L$, the original quantity is $10^L$. Here, $\text{pH} = -\log_{10}[\text{H}^+]$, so rearrange to find $[\text{H}^+]$.

   1. Write the equation with the known pH: $3 = -\log_{10}[\text{H}^+]$
   2. Multiply both sides by $-1$: $\log_{10}[\text{H}^+] = -3$
   3. Convert from logarithmic form to index form: $[\text{H}^+] = 10^{-3}$
   4. Evaluate: $[\text{H}^+] = 0.001 \text{ mol/L}$

   Answer: The hydrogen ion concentration is $0.001$ mol/L.

8. Laws of logarithms

   Solve for $x$, checking that your answer is valid:\

   $$\log_3(x + 6) + \log_3 x = 3$$
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   Key idea: Use the product law $\log_a m + \log_a n = \log_a(mn)$ to combine the left side, then convert to exponential form to solve.

   1. Apply the product law:\

   $$\log_3\bigl(x(x+6)\bigr) = 3$$

   2. Convert to exponential form:\

   $$x(x + 6) = 3^3 = 27$$

   3. Expand and rearrange:\

   $$x^2 + 6x - 27 = 0$$

   4. Factorise:\

   $$(x + 9)(x - 3) = 0 \implies x = -9 \text{ or } x = 3$$

   5. Validity check: both $\log_3(x+6)$ and $\log_3 x$ require their arguments to be positive, so $x > 0$. Reject $x = -9$.

   Answer: $x = 3$

Free worksheets for this chapter

Worked examples and a free PDF with answers for each topic.

• Year 10 index laws with algebra worksheet
• Year 10 logarithmic scales worksheet
• Year 10A surds and fractional indices worksheet
• Year 10 exponential equations worksheet
• Year 10A exponential and logarithmic equations worksheet
• Year 10 graphs of parabolas, circles, hyperbolas and exponentials worksheet
• Year 10 compound interest worksheet
• Year 10A logarithms and log laws worksheet