Indices, Surds & Logarithms Practice

Q: When do I need to change everything to a common base?

Whenever an index equation has different bases, such as 8x=4x+18^{x}=4^{x+1}8x=4x+1. Rewrite each base as a power of the same number, here 222, then equate the indices. The same idea powers many logarithm questions too.

Q: Why must I reject one answer in the last question?

Because log⁡2x\log_{2}xlog2x is only defined when x>0x>0x>0. The quadratic gives x=4x=4x=4 or x=−2x=-2x=−2, but x=−2x=-2x=−2 would need the logarithm of a negative number, so it is not valid. Always test both roots in the original equation.

How to use these practice questions

The six questions below rise in difficulty across the whole chapter, from a single law of indices to solving a logarithmic equation with a root to reject. Give yourself roughly five to eight minutes per question and work on paper first, writing every line the way you would in the real exam, a method line that names the law you are using, a clear substitution, then the final answer.

Resist the urge to peek.

Only once you have committed to a full answer should you open the solution and mark yourself line by line. When your working differs from ours, stop at the exact line where the two part company; that single step is almost always where the mark was lost.

Because Add Math is marked analytically, a correct law stated clearly still earns credit even when the arithmetic slips, so always show your steps in full.

Six practice questions

Q1[3 marks]

Simplify $\dfrac{(2x^{2})^{3}}{4x}$ , giving your answer in the form $kx^{n}$ .

Show worked solution

Deal with the power of a product on top first: raise both the number and the letter to the power $3$ . Using $(ab)^{n}=a^{n}b^{n}$ and $(x^{m})^{n}=x^{mn}$ :

(2x^{2})^{3}=2^{3}\,x^{2\times 3}=8x^{6}

Now divide by $4x$ . Divide the numbers, and for the letters subtract the indices using $x^{a}\div x^{b}=x^{a-b}$ :

\frac{8x^{6}}{4x}=\frac{8}{4}\,x^{6-1}=2x^{5}

Answer

$2x^{5}$ . Check with a value: at $x=1$ , the original is $\dfrac{(2)^{3}}{4}=\dfrac{8}{4}=2$ , and $2(1)^{5}=2$ , so the two agree.

Q2[3 marks]

Solve the equation $8^{x}=4^{x+1}$ .

Show worked solution

The bases are different, so first write both $8$ and $4$ as powers of the same base $2$ . Since $8=2^{3}$ and $4=2^{2}$ :

8^{x}=(2^{3})^{x}=2^{3x} \qquad 4^{x+1}=(2^{2})^{x+1}=2^{2x+2}

With equal bases, the indices must be equal. Set them equal and solve the linear equation:

3x=2x+2 \;\Rightarrow\; x=2

Answer

$x=2$ . Check by substituting back: $8^{2}=64$ and $4^{2+1}=4^{3}=64$ , so both sides equal $64$ .

Q3[4 marks]

(a) Simplify $\sqrt{50}+\sqrt{18}$ , giving your answer in the form $a\sqrt{2}$ . (b) Rationalise the denominator of $\dfrac{6}{\sqrt{3}}$ .

Show worked solution

(a) Simplify each surd by taking out the largest perfect square factor, then collect like surds. Since $50=25\times 2$ and $18=9\times 2$ :

\sqrt{50}=\sqrt{25}\,\sqrt{2}=5\sqrt{2} \qquad \sqrt{18}=\sqrt{9}\,\sqrt{2}=3\sqrt{2}

\sqrt{50}+\sqrt{18}=5\sqrt{2}+3\sqrt{2}=8\sqrt{2}

(b) Multiply the top and bottom by $\sqrt{3}$ so the denominator becomes rational, using $\sqrt{3}\times\sqrt{3}=3$ :

\frac{6}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{6\sqrt{3}}{3}=2\sqrt{3}

Answer

(a) $8\sqrt{2}$ ; (b) $2\sqrt{3}$ . Decimal check: $\sqrt{50}+\sqrt{18}\approx 7.07+4.24=11.31$ and $8\sqrt{2}\approx 11.31$ ; also $\dfrac{6}{\sqrt{3}}\approx 3.46$ and $2\sqrt{3}\approx 3.46$ .

Q4[4 marks]

Express $\dfrac{5}{\sqrt{7}+\sqrt{2}}$ with a rational denominator, giving your answer in simplest surd form.

Show worked solution

When the denominator is a sum of two surds, multiply top and bottom by its conjugate, the same two surds with the sign between them changed. Here the conjugate of $\sqrt{7}+\sqrt{2}$ is $\sqrt{7}-\sqrt{2}$ :

\frac{5}{\sqrt{7}+\sqrt{2}}\times\frac{\sqrt{7}-\sqrt{2}}{\sqrt{7}-\sqrt{2}}

The denominator is now a difference of two squares, $(\sqrt{7})^{2}-(\sqrt{2})^{2}$ , which removes the surds:

=\frac{5(\sqrt{7}-\sqrt{2})}{7-2}=\frac{5(\sqrt{7}-\sqrt{2})}{5}=\sqrt{7}-\sqrt{2}

Answer

$\sqrt{7}-\sqrt{2}$ . Decimal check: $\sqrt{7}-\sqrt{2}\approx 2.646-1.414=1.232$ , and $\dfrac{5}{\sqrt{7}+\sqrt{2}}\approx\dfrac{5}{4.060}=1.232$ , which agrees.

Q5[4 marks]

Given that $\log_{a}2=m$ and $\log_{a}3=n$ , express in terms of $m$ and $n$ : (a) $\log_{a}18$ , and (b) $\log_{a}\dfrac{4}{3}$ .

Show worked solution

(a) Break $18$ into factors that are powers of $2$ and $3$ : $18=2\times 3^{2}$ . Then use $\log(xy)=\log x+\log y$ and $\log x^{k}=k\log x$ :

\log_{a}18=\log_{a}2+\log_{a}3^{2}=\log_{a}2+2\log_{a}3=m+2n

(b) Use the quotient law $\log\dfrac{x}{y}=\log x-\log y$ , and write $4=2^{2}$ :

\log_{a}\frac{4}{3}=\log_{a}4-\log_{a}3=2\log_{a}2-\log_{a}3=2m-n

Answer

(a) $m+2n$ ; (b) $2m-n$ . The key move is factorising the number inside the logarithm into $2$ s and $3$ s, because only those two logarithms are given.

Q6[5 marks]

Solve the equation $\log_{2}x+\log_{2}(x-2)=3$ .

Show worked solution

The two logarithms have the same base, so combine them with the product law $\log_{2}A+\log_{2}B=\log_{2}(AB)$ :

\log_{2}\big[x(x-2)\big]=3

Rewrite in index form: $\log_{2}N=3$ means $N=2^{3}=8$ . This clears the logarithm:

x(x-2)=2^{3}=8 \;\Rightarrow\; x^{2}-2x-8=0

Factorise and solve the quadratic:

(x-4)(x+2)=0 \;\Rightarrow\; x=4 \;\text{or}\; x=-2

A logarithm is only defined for a positive number, so $x$ and $x-2$ must both be positive. That rejects $x=-2$ , leaving $x=4$ .

Answer

$x=4$ . Check: $\log_{2}4+\log_{2}2=2+1=3$ , as required.

Always test each root in the original equation, the negative one gives the logarithm of a negative number, so it cannot be a valid answer.

How to mark yourself like an examiner

Add Math is marked analytically, which means marks are attached to steps, not only to the final number. When you check your own script, award yourself credit the way a marker would: look for the correct law being used, the right substitution, and a clean final statement.

Method mark: did you name and use the correct law, a law of indices, taking out a perfect square, multiplying by the conjugate, or a law of logarithms?
Substitution mark: for an index equation, are both sides written to a common base before you equate the powers?
Answer mark: is the final value or surd stated in simplest form, and does it survive a check by substituting back?
For a logarithmic equation, you only earn full marks if you reject any root that makes the number inside a logarithm zero or negative.
If your final number is wrong but the law and substitution are right, give yourself the method marks, that is exactly what a real marker does.

How a teacher helps

Marking yourself is powerful, but it is hard to see your own blind spots. In a one-to-one lesson our teacher watches the exact line where a mark slips away, a base not converted, a surd left un-simplified, or a rejected root forgotten in a log equation, and corrects the habit on the spot.

Because our teachers are experienced, you work with someone who explains the why behind each law. Lessons are taught in English, while SPM papers are set in both Malay and English, so we make sure the notation reads the same to you either way.

Get 1-to-1 help.

Book a Trial Class

Frequently asked questions

How long should each of these questions take me?

Aim for roughly five to eight minutes each, rising with the mark value. If a question takes far longer, note it and bring it to a lesson, the time it steals in the exam is often the real problem, not the topic itself.

When do I need to change everything to a common base?

Whenever an index equation has different bases, such as $8^{x}=4^{x+1}$ . Rewrite each base as a power of the same number, here $2$ , then equate the indices.

The same idea powers many logarithm questions too.

Why must I reject one answer in the last question?

Because $\log_{2}x$ is only defined when $x>0$ . The quadratic gives $x=4$ or $x=-2$ , but $x=-2$ would need the logarithm of a negative number, so it is not valid.

Always test both roots in the original equation.

Do I lose all the marks if my final answer is wrong?

No. Because marking is analytic, a correct law and a correct substitution still earn marks even if the arithmetic slips at the end.

That is why you should always show full working.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Indices, Surds and Logarithms, Practice Questions

How to use these practice questions

Six practice questions

How to mark yourself like an examiner

How a teacher helps

Frequently asked questions

Ready to get started?