Indices and Powers

The Laws of Indices

In $a^n$, $a$ is the base and $n$ is the index (also called the exponent or the power). Six laws do almost all the work in this topic.

The first three need the same base. The last three need the same index. When neither matches, as in $2^3 \times 3^4$, nothing combines until you rewrite the numbers.

The traps.

• Adding is not multiplying. $2^5 + 2^5 = 2 \times 2^5 = 2^6$, not $2^{10}$. In the same way $3^4 + 3^4 + 3^4 = 3 \times 3^4 = 3^5$. Adding $a$ copies of $a^n$ gives $a^{n+1}$.
• Towers are read from the top. $\left(2^3\right)^2 = 2^6 = 64$, but $2^{3^2} = 2^9 = 512$.
• A power does not spread over a sum. $(2 + 3)^2 = 25$, not $2^2 + 3^2 = 13$.
• The minus sign. $-3^2 = -9$, because the power belongs only to the 3, while $(-3)^2 = 9$. An even power of a negative number is positive and an odd power is negative: $(-2)^6 = 64$ and $(-2)^5 = -32$. So $(-1)^n$ is 1 for even $n$ and −1 for odd $n$.

Taking out a common power. The laws work across × and ÷ only. With + or −, take the smallest power out first:

$$2^{10} + 2^{11} + 2^{12} = 2^{10}(1 + 2 + 4) = 7 \times 2^{10} = 7168$$

Zero, Negative and Fractional Indices

Zero index. $a^0 = 1$ for any $a \neq 0$. It comes straight from the division law: $\frac{a^n}{a^n} = a^{n-n} = a^0$, and anything divided by itself is 1. So $7^0 = 1$, $(-5)^0 = 1$ and $\left(\frac{2}{3}\right)^0 = 1$. Watch the brackets: $3x^0 = 3$, but $(3x)^0 = 1$.

Negative index. $a^{-n} = \frac{1}{a^n}$.

• $2^{-3} = \frac{1}{8}$ and $10^{-2} = 0.01$.
• $\left(\frac{2}{5}\right)^{-2} = \left(\frac{5}{2}\right)^2 = \frac{25}{4}$. A negative index flips the fraction.
• $\frac{3^{-2}}{2^{-3}} = \frac{2^3}{3^2} = \frac{8}{9}$. Moving a power across the fraction line changes the sign of its index.

A negative index never makes the number negative. $2^{-3}$ is a small positive number, not −8.

Fractional index. $a^{\frac{1}{n}} = \sqrt[n]{a}$ and $a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m$. The bottom of the fraction is the root and the top is the power. Take the root first so that the numbers stay small.

• $8^{\frac{2}{3}} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4$
• $32^{\frac{3}{5}} = \left(\sqrt[5]{32}\right)^3 = 2^3 = 8$
• $81^{-\frac{3}{4}} = \frac{1}{\left(\sqrt[4]{81}\right)^3} = \frac{1}{27}$
• $\left(\frac{27}{125}\right)^{-\frac{2}{3}} = \left(\frac{125}{27}\right)^{\frac{2}{3}} = \left(\frac{5}{3}\right)^2 = \frac{25}{9}$

For $a^{-\frac{m}{n}}$ the order of moves is: flip (for the minus), root (for $n$), power (for $m$).

Decimal indices. Change them into fractions: $16^{0.75} = 16^{\frac{3}{4}} = 8$, $32^{0.2} = 32^{\frac{1}{5}} = 2$ and $9^{1.5} = 9^{\frac{3}{2}} = 27$.

Negative bases. An odd root of a negative number is fine: $(-8)^{\frac{1}{3}} = -2$ and $(-32)^{\frac{2}{5}} = (-2)^2 = 4$. An even root of a negative number is not a real number, so it never turns up as a correct answer.

Start with a prime base. Write the base as a power of a small prime before you do anything else: $0.25 = 2^{-2}$, $\frac{1}{125} = 5^{-3}$ and $0.0016 = 0.2^4$. Then the indices just multiply. For example, $\left(\frac{1}{125}\right)^{-\frac{2}{3}} = \left(5^{-3}\right)^{-\frac{2}{3}} = 5^2 = 25$.

Surds Written as Indices

Every root can be written as a fractional index, and once it is, all the laws apply.

$$\sqrt[n]{a} = a^{\frac{1}{n}} \qquad \sqrt[n]{a^m} = a^{\frac{m}{n}} \qquad \frac{1}{\sqrt{a}} = a^{-\frac{1}{2}}$$

• $\sqrt{x} = x^{\frac{1}{2}}$, $\sqrt[3]{x^2} = x^{\frac{2}{3}}$ and $\sqrt[5]{x^{10}} = x^2$.
• A root of a root multiplies the orders: $\sqrt[3]{\sqrt{x}} = x^{\frac{1}{6}}$, and in general $\sqrt[m]{\sqrt[n]{x}} = \sqrt[mn]{x}$.
• A root of a product: $\sqrt{x\sqrt[3]{x}} = \left(x^{\frac{4}{3}}\right)^{\frac{1}{2}} = x^{\frac{2}{3}}$.

Multiplying roots with the same base. Add the indices: $\sqrt{2} \times \sqrt[3]{2} = 2^{\frac{1}{2} + \frac{1}{3}} = 2^{\frac{5}{6}} = \sqrt[6]{32}$. Dividing subtracts them: $\sqrt[3]{4} \div \sqrt{2} = 2^{\frac{2}{3} - \frac{1}{2}} = 2^{\frac{1}{6}} = \sqrt[6]{2}$.

Multiplying roots with different bases. Bring them to the same order, using the LCM of the orders:

$$\sqrt{2} \times \sqrt[3]{3} = 2^{\frac{3}{6}} \times 3^{\frac{2}{6}} = \sqrt[6]{2^3 \times 3^2} = \sqrt[6]{72}$$

Changing the order of a root. Raising both the number and the order by the same factor changes nothing: $\sqrt{3} = \sqrt[4]{9} = \sqrt[6]{27}$. This is the key step for comparing roots later in this section.

Same order, different bases. $a^{\frac{1}{n}} \times b^{\frac{1}{n}} = (ab)^{\frac{1}{n}}$, so $\sqrt{8} \times \sqrt{18} = \sqrt{144}$ = 12 and $\sqrt[4]{27} \times \sqrt[4]{3} = \sqrt[4]{81}$ = 3.

Repeated roots. $\sqrt{x\sqrt{x\sqrt{x}}} = x^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8}} = x^{\frac{7}{8}}$, because the outer $x$ is under one root, the next under two and the last under three. The Squares, Square Roots and Cube Roots section has the full list of nested and infinite forms.

[SSC Tip] Options often mix the two ways of writing the same thing. $\sqrt[6]{32}$, $2^{\frac{5}{6}}$ and $\sqrt{2} \times \sqrt[3]{2}$ are all one number, so turn every option into a power of a prime before you compare.

Simplifying Power Expressions

Step 1: break every base into primes. Then collect the powers of each prime: add the indices on top, subtract the ones underneath.

$$\frac{15^3 \times 6^4}{10^3 \times 9^4} = \frac{\left(3^3 \times 5^3\right) \times \left(2^4 \times 3^4\right)}{\left(2^3 \times 5^3\right) \times 3^8}$$

So the value is $2 \times 3^{-1} = \frac{2}{3}$.

Step 2: with + or −, take out the smallest power. The laws only work across × and ÷, so a sum has to be factorised first.

$$\frac{2^{n+4} - 2 \times 2^n}{2 \times 2^{n+3}} = \frac{2^n(16 - 2)}{2^n \times 16} = \frac{14}{16} = \frac{7}{8}$$

$$\frac{5^{n+2} - 6 \times 5^{n+1}}{13 \times 5^n - 2 \times 5^{n+1}} = \frac{5^n(25 - 30)}{5^n(13 - 10)} = -\frac{5}{3}$$

The $n$ always cancels in these, so a quick check is to put $n$ = 0 and work out plain numbers. With $n$ = 0 the first one is $\frac{16 - 2}{2 \times 8} = \frac{7}{8}$.

Step 3: decimals and fractions as bases. Turn a decimal into a fraction or a power: $(0.25)^{-1.5} = \left(\frac{1}{4}\right)^{-\frac{3}{2}} = 4^{\frac{3}{2}} = 8$.

Letters work the same way.

$$\left(a^2 b^{-3}\right)^{-2} \times \left(a^{-1} b^2\right)^3 = a^{-4} b^6 \times a^{-3} b^6 = a^{-7} b^{12} = \frac{b^{12}}{a^7}$$

Negative indices inside a sum. $\left(2^{-1} + 3^{-1}\right)^{-1} = \left(\frac{1}{2} + \frac{1}{3}\right)^{-1} = \left(\frac{5}{6}\right)^{-1} = \frac{6}{5}$. It is not 2 + 3 = 5: the outer index flips the whole sum, not each term.

Solving Exponential Equations

Same base: equate the indices. If $a^x = a^y$ (with $a > 0$ and $a \neq 1$), then $x = y$. So the whole job is to write both sides as powers of one base.

• $2^{x+3} = 32 = 2^5$, so $x + 3 = 5$ and $x$ = 2.
• $4^{x+1} = 8^{x-1}$ becomes $2^{2x+2} = 2^{3x-3}$, so $2x + 2 = 3x - 3$ and $x$ = 5.
• $\left(\sqrt{3}\right)^x = 81$ becomes $3^{\frac{x}{2}} = 3^4$, so $x$ = 8.
• $2^x \times 8^{\frac{1}{5}} = 2^{\frac{1}{5}}$ becomes $2^{x + \frac{3}{5}} = 2^{\frac{1}{5}}$, so $x = -\frac{2}{5}$.

Reciprocal bases. Flip one side and change the sign of its index.

• $\left(\frac{3}{4}\right)^x = \frac{64}{27} = \left(\frac{4}{3}\right)^3 = \left(\frac{3}{4}\right)^{-3}$, so $x$ = −3.
• $\left(\frac{a}{b}\right)^{x-1} = \left(\frac{b}{a}\right)^{x-3} = \left(\frac{a}{b}\right)^{3-x}$, so $x - 1 = 3 - x$ and $x$ = 2.

Equal to 1. $a^x = 1$ means $x = 0$ (for $a \neq 1$). So $7^{x-2} = 1$ gives $x$ = 2. In the same way, two different bases with the same index can only be equal when that index is 0: $2^{x-3} = 5^{x-3}$ gives $x$ = 3.

Sums of powers: take out the smallest.

• $2^x + 2^{x+1} + 2^{x+2} = 112$: $2^x(1 + 2 + 4) = 112$, so $2^x$ = 16 and $x$ = 4.
• $4^x - 4^{x-1} = 24$: $4^{x-1}(4 - 1) = 24$, so $4^{x-1}$ = 8. As powers of 2, $2^{2x-2} = 2^3$, so $x = \frac{5}{2}$.

[Tier 2] A quadratic in disguise. In $9^x - 4 \times 3^x + 3 = 0$, put $t = 3^x$, so that $9^x = t^2$. Then $t^2 - 4t + 3 = 0$ gives $t$ = 1 or 3, so $x$ = 0 or 1.

[SSC Tip] The options are almost always small numbers. If the algebra looks long, put each option into the equation; the right one usually shows itself within two tries.

Powers of 10, Decimals and Counting Digits

Decimals as powers of 10. 0.1 = $10^{-1}$, 0.01 = $10^{-2}$ and 0.001 = $10^{-3}$: the number of decimal places is the negative index. So $(0.001)^2 \times 10^5 = 10^{-6} \times 10^5 = 10^{-1}$ = 0.1.

Decimals as powers of a smaller decimal. Spot the power hidden in the digits, then count the places.

• 0.0081 = $0.3^4$, since $3^4$ = 81 and there are four places. So $(0.0081)^{\frac{1}{4}}$ = 0.3 and $(0.0081)^{-\frac{1}{4}} = \frac{10}{3}$.
• 0.008 = $0.2^3$, so $(0.008)^{\frac{2}{3}} = 0.2^2$ = 0.04.
• 0.0036 = $0.06^2$, so $(0.0036)^{-\frac{1}{2}} = \frac{1}{0.06} = \frac{50}{3}$.

Decimal indices on the same base. SSC hides a simple fraction inside decimals that add up nicely: $243^{0.12} \times 243^{0.08} = 243^{0.2} = 243^{\frac{1}{5}}$ = 3. In the same way $(6.25)^{0.5}$ = 2.5 and $(1.44)^{1.5} = 1.2^3$ = 1.728.

Scientific form. This means writing a number as $a \times 10^n$ with $1 \le a < 10$: 0.000045 = $4.5 \times 10^{-5}$ and 3,20,000 = $3.2 \times 10^5$. To multiply, multiply the front parts and add the indices: $\left(4 \times 10^5\right) \times \left(6 \times 10^{-8}\right) = 24 \times 10^{-3}$ = 0.024.

How many digits? Pair each 2 with a 5 to make a 10. Whatever is left over sits in front of the zeros.

• $4^{11} \times 5^{25} = 2^{22} \times 5^{25} = 5^3 \times 10^{22} = 125 \times 10^{22}$. That is 125 followed by 22 zeros, so 25 digits.
• $8^5 \times 5^{12} = 2^{15} \times 5^{12} = 2^3 \times 10^{12} = 8 \times 10^{12}$, so 13 digits.

[Tier 2] Rough sizes. $2^{10}$ = 1024 is just over $10^3$. So $2^{40}$ is a little over $10^{12}$ (about $1.1 \times 10^{12}$) and has 13 digits, and $2^{100}$ is a little over $10^{30}$, so it has 31 digits.

Don't mix up "zeros at the end" and "number of digits". $125 \times 10^{22}$ ends in 22 zeros but has 25 digits.

Comparing Large Powers

Nobody works out $3^{200}$. You make either the bases or the indices match, and then compare the part that is different.

Make the bases equal. Use prime powers. $4^{50} = 2^{100}$, $8^{34} = 2^{102}$ and $16^{25} = 2^{100}$. With the same base (bigger than 1), the bigger index wins, so $8^{34}$ is the largest, and $4^{50} = 16^{25}$.

Make the indices equal. Use the HCF of the indices. For $2^{300}$ and $3^{200}$ the HCF is 100:

$$2^{300} = \left(2^3\right)^{100} = 8^{100} \qquad 3^{200} = \left(3^2\right)^{100} = 9^{100}$$

With the same index, the bigger base wins, so $3^{200}$ is larger.

The same idea puts a whole list in order. For $2^{60}$, $3^{40}$, $5^{30}$ and $7^{20}$ the HCF is 10, which gives $64^{10}$, $81^{10}$, $125^{10}$ and $49^{10}$. So $7^{20} < 2^{60} < 3^{40} < 5^{30}$.

Big base or big index? Take $100^{50}$ against $50^{100}$. Write $50^{100} = \left(50^2\right)^{50} = 2500^{50}$. Now both have index 50, and 2500 beats 100, so $50^{100}$ is far bigger. Doubling the index usually matters more than doubling the base.

Bases between 0 and 1. The rule flips: a bigger index makes the number smaller. $(0.5)^3 = 0.125$ is less than $(0.5)^2 = 0.25$.

[Tier 2] When nothing matches, squeeze. Compare each number with a power of 2 next to it. For $31^{11}$ and $17^{14}$: $31^{11} < 32^{11} = 2^{55}$, and $17^{14} > 16^{14} = 2^{56}$. So $31^{11} < 2^{55} < 2^{56} < 17^{14}$.

In "which is the largest" questions the HCF of the indices usually comes out as 10, 12, 50 or 100, and the bases you get are small enough to compare at a glance.

Comparing Roots of Different Orders

To compare $\sqrt{2}$, $\sqrt[3]{3}$, $\sqrt[4]{4}$ and $\sqrt[6]{6}$, raise every one to the LCM of the orders. Raising positive numbers to the same positive power keeps them in the same order, and it clears all the roots. Here the LCM of 2, 3, 4 and 6 is 12.

So $\sqrt[3]{3}$ is the largest, $\sqrt[6]{6}$ is the smallest, and $\sqrt{2}$ and $\sqrt[4]{4}$ are equal (as they must be, since $\sqrt[4]{4} = \sqrt[4]{2^2} = \sqrt{2}$).

The $\sqrt[n]{n}$ pattern. When the number and the order are the same, the value peaks at $n$ = 3 and then falls: $\sqrt[3]{3} > \sqrt{2} = \sqrt[4]{4} > \sqrt[5]{5} > \sqrt[6]{6} > \cdots$. This only holds when the number matches the order. With other numbers, work it out: $\sqrt[6]{12}$ (12th power 144) beats $\sqrt[3]{3}$ (12th power 81), even though its order is bigger.

Another mix. For $\sqrt[3]{4}$, $\sqrt{3}$ and $\sqrt[6]{10}$ the LCM is 6. The 6th powers are $4^2$ = 16, $3^3$ = 27 and $10^1$ = 10, so $\sqrt{3} > \sqrt[3]{4} > \sqrt[6]{10}$.

Same number, different orders. For a number bigger than 1, a higher-order root is smaller: $\sqrt{5} > \sqrt[3]{5} > \sqrt[4]{5}$. For a number between 0 and 1 it is the other way round: $\sqrt{0.5} < \sqrt[3]{0.5}$.

Same order, different numbers. The bigger number has the bigger root: $\sqrt[3]{7} > \sqrt[3]{6}$.

Comparing sums and differences of roots, such as $\sqrt{5} + \sqrt{3}$ against $\sqrt{6} + \sqrt{2}$, needs squaring or rationalising, and it is in the Surds and Rationalisation section.

[SSC Tip] If the LCM makes the powers big, compare two at a time. $\sqrt[4]{10}$ against $\sqrt[3]{6}$ needs only the 12th powers $10^3$ = 1000 and $6^4$ = 1296, so $\sqrt[3]{6}$ is bigger.

Index Identities with Letters [Tier 2]

These show up mainly in Tier 2. Each one is proved by adding indices, so you don't need to memorise them if you can do that step.

Cyclic indices cancel to zero.

$$\left(x^{a-b}\right)^{a+b} \times \left(x^{b-c}\right)^{b+c} \times \left(x^{c-a}\right)^{c+a} = x^{(a^2 - b^2) + (b^2 - c^2) + (c^2 - a^2)} = x^0 = 1$$

In the same way, $\left(\frac{x^b}{x^c}\right)^a \times \left(\frac{x^c}{x^a}\right)^b \times \left(\frac{x^a}{x^b}\right)^c = x^{(ab - ac) + (bc - ab) + (ac - bc)} = 1$.

Reciprocal pairs add to 1.

$$\frac{1}{1 + x^{a-b}} + \frac{1}{1 + x^{b-a}} = 1$$

Multiply the top and bottom of the second fraction by $x^{a-b}$. It becomes $\frac{x^{a-b}}{x^{a-b} + 1}$, and the two fractions add to 1. The three-term version, $\frac{1}{1 + x^{b-a} + x^{c-a}} + \frac{1}{1 + x^{a-b} + x^{c-b}} + \frac{1}{1 + x^{a-c} + x^{b-c}}$, is also 1: multiplying the first fraction through by $x^a$ gives $\frac{x^a}{x^a + x^b + x^c}$, and the other two turn into $x^b$ and $x^c$ over the same bottom.

Equal powers: $a^x = b^y = c^z$. Call the common value $k$. Then $a = k^{\frac{1}{x}}$, $b = k^{\frac{1}{y}}$ and $c = k^{\frac{1}{z}}$, and any link between $a$, $b$ and $c$ turns into a link between $\frac{1}{x}$, $\frac{1}{y}$ and $\frac{1}{z}$.

• If $abc = 1$, then $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$, which is the same as $xy + yz + zx = 0$.
• If $b^2 = ac$, then $\frac{2}{y} = \frac{1}{x} + \frac{1}{z}$.
• If $2^x = 3^y = 6^z$, then since 6 = 2 × 3, $\frac{1}{z} = \frac{1}{x} + \frac{1}{y}$.
• If $2^x = 3^y = 12^z$, then since 12 = $2^2 \times 3$, $\frac{1}{z} = \frac{2}{x} + \frac{1}{y}$.

Chains. If $a = b^x$, $b = c^y$ and $c = a^z$, then $a = b^x = c^{xy} = a^{xyz}$, so $xyz = 1$.

[SSC Tip] These hold for every value, so you can test the options with easy numbers. For $2^x = 4^y = 8^z$, set each one equal to 64: then $x$ = 6, $y$ = 3 and $z$ = 2, and a claim like $\frac{1}{x} + \frac{1}{z} = \frac{2}{y}$ can be checked in your head ($\frac{1}{6} + \frac{1}{2} = \frac{2}{3}$, which matches).

Solved Examples

Question 1

Simplify $\frac{16^3 \times 8^2}{32^4}$.

Answer. Everything is a power of 2: $16^3 = 2^{12}$, $8^2 = 2^6$ and $32^4 = 2^{20}$. So the value is $\frac{2^{12 + 6}}{2^{20}} = 2^{-2}$, which is $\frac{1}{4}$.

Question 2

Find $64^{-\frac{2}{3}} \times 16^{\frac{3}{4}}$.

Answer. Root first, then power. $\sqrt[3]{64}$ = 4, so $64^{\frac{2}{3}}$ = 16 and $64^{-\frac{2}{3}} = \frac{1}{16}$. $\sqrt[4]{16}$ = 2, so $16^{\frac{3}{4}}$ = 8. The product is $\frac{8}{16}$, which is $\frac{1}{2}$.

Question 3

Find $\left(\frac{27}{64}\right)^{-\frac{2}{3}}$.

Answer. The minus sign flips the fraction, giving $\left(\frac{64}{27}\right)^{\frac{2}{3}}$. The cube root is $\frac{4}{3}$, and squaring gives $\frac{16}{9}$.

Question 4

Find $(-27)^{\frac{2}{3}}$ and $(-32)^{-\frac{3}{5}}$.

Answer. Odd roots of negative numbers are fine. $\sqrt[3]{-27} = -3$, and $(-3)^2$ = 9. For the second, $\sqrt[5]{-32} = -2$, then $(-2)^3 = -8$, and the negative index flips it to $-\frac{1}{8}$.

Question 5

Simplify $\frac{12^4 \times 9^3 \times 4}{6^3 \times 8^2 \times 27}$.

Answer. I write each number in primes.

• Top: $12^4 = 2^8 \times 3^4$, $9^3 = 3^6$ and $4 = 2^2$. Together, $2^{10} \times 3^{10}$.
• Bottom: $6^3 = 2^3 \times 3^3$, $8^2 = 2^6$ and $27 = 3^3$. Together, $2^9 \times 3^6$.

Dividing leaves $2^1 \times 3^4$ = 2 × 81 = 162.

Question 6

Simplify $\frac{3^{n+3} - 6 \times 3^{n+1}}{3^{n+2} + 3^{n+1}}$.

Answer. I take $3^{n+1}$ out of the top and the bottom. Top: $3^{n+1}(9 - 6) = 3 \times 3^{n+1}$. Bottom: $3^{n+1}(3 + 1) = 4 \times 3^{n+1}$. The $3^{n+1}$ cancels, leaving $\frac{3}{4}$. A quick check with $n$ = 0: $\frac{27 - 18}{9 + 3} = \frac{9}{12} = \frac{3}{4}$.

Question 7

Find $\sqrt[3]{2} \times \sqrt[4]{2} \times \sqrt[12]{32}$.

Answer. As powers of 2 this is $2^{\frac{1}{3}} \times 2^{\frac{1}{4}} \times 2^{\frac{5}{12}}$. With 12 as the common denominator, the indices add up to $\frac{4}{12} + \frac{3}{12} + \frac{5}{12} = 1$. So the value is $2^1$ = 2.

Question 8

Write $\sqrt[4]{\sqrt[3]{x^2}}$ as a power of $x$, and find its value when $x$ = 64.

Answer. The inside is $x^{\frac{2}{3}}$, and the fourth root multiplies the index by $\frac{1}{4}$, giving $x^{\frac{2}{12}} = x^{\frac{1}{6}}$. When $x$ = 64 = $2^6$, the value is $2^1$ = 2.

Question 9

Simplify $\sqrt{x\sqrt[3]{x\sqrt{x}}}$.

Answer. I work from the inside. $x\sqrt{x} = x^{\frac{3}{2}}$, and its cube root is $x^{\frac{1}{2}}$. Then $x \times x^{\frac{1}{2}} = x^{\frac{3}{2}}$, and the outer square root gives $x^{\frac{3}{4}}$.

Question 10

Solve $25^{x-1} = 125^{x-3}$.

Answer. Both are powers of 5: $5^{2x-2} = 5^{3x-9}$. The bases match, so the indices must match: $2x - 2 = 3x - 9$, which gives $x$ = 7.

Question 11

If $\left(\frac{3}{5}\right)^x \times \left(\frac{5}{3}\right)^{2x} = \frac{125}{27}$, find $x$.

Answer. I flip the first fraction: $\left(\frac{3}{5}\right)^x = \left(\frac{5}{3}\right)^{-x}$. The left side becomes $\left(\frac{5}{3}\right)^{-x + 2x} = \left(\frac{5}{3}\right)^x$, and the right side is $\left(\frac{5}{3}\right)^3$. So $x$ = 3.

Question 12

If $3^{x+2} + 3^x = 270$, find $x$.

Answer. I take out the smaller power: $3^x\left(3^2 + 1\right) = 270$, so $10 \times 3^x = 270$ and $3^x$ = 27 = $3^3$. So $x$ = 3.

Question 13

If $\left(\sqrt{2}\right)^7 \times 4^3 = 2^n \times 2\sqrt{2}$, find $n$.

Answer. Left side: $\left(\sqrt{2}\right)^7 = 2^{\frac{7}{2}}$ and $4^3 = 2^6$, so together they make $2^{\frac{19}{2}}$. Right side: $2\sqrt{2} = 2^{\frac{3}{2}}$, so it is $2^{n + \frac{3}{2}}$. Then $n + \frac{3}{2} = \frac{19}{2}$, and $n$ = 8.

Question 14

Solve $5^{2x-1} - 25^{x-1} = 2500$.

Answer. I write both terms using $25^x$, since $5^{2x} = 25^x$: $5^{2x-1} = \frac{25^x}{5}$ and $25^{x-1} = \frac{25^x}{25}$. So $25^x\left(\frac{1}{5} - \frac{1}{25}\right) = 2500$, which is $25^x \times \frac{4}{25} = 2500$. That gives $25^x$ = 15625 = $25^3$, so $x$ = 3.

Question 15 [Tier 2]

Solve $4^x - 3 \times 2^{x+1} + 8 = 0$.

Answer. I put $t = 2^x$. Then $4^x = t^2$ and $2^{x+1} = 2t$, so the equation is $t^2 - 6t + 8 = 0$, or $(t - 2)(t - 4) = 0$. So $2^x$ = 2 or 4, which means $x$ = 1 or $x$ = 2.

Question 16

Find $(0.0016)^{-\frac{3}{4}}$.

Answer. 0.0016 = $0.2^4$, since $2^4$ = 16 and there are four decimal places. So $(0.0016)^{-\frac{3}{4}} = (0.2)^{-3} = \frac{1}{0.008}$ = 125.

Question 17

How many digits are there in $8^9 \times 5^{25}$?

Answer. $8^9 = 2^{27}$. I pair 25 of the 2s with the 25 fives to make $10^{25}$, and $2^2$ = 4 is left over. So the number is $4 \times 10^{25}$: a 4 followed by 25 zeros, which is 26 digits.

Question 18

Which is the largest: $2^{250}$, $3^{150}$ or $5^{100}$?

Answer. The indices are all multiples of 50. $2^{250} = \left(2^5\right)^{50} = 32^{50}$, $3^{150} = \left(3^3\right)^{50} = 27^{50}$ and $5^{100} = \left(5^2\right)^{50} = 25^{50}$. With the same index the biggest base wins, so $2^{250}$ is the largest.

Question 19

Arrange $\sqrt[3]{5}$, $\sqrt{3}$ and $\sqrt[4]{7}$ in descending order.

Answer. The LCM of 3, 2 and 4 is 12, so I raise each one to the 12th power: $5^4$ = 625, $3^6$ = 729 and $7^3$ = 343. In descending order that is $\sqrt{3} > \sqrt[3]{5} > \sqrt[4]{7}$.

Question 20 [Tier 2]

Which is greater, $31^{12}$ or $17^{15}$?

Answer. The bases and indices don't line up, so I compare both with a power of 2. $31^{12} < 32^{12} = 2^{60}$, and $17^{15} > 16^{15} = 2^{60}$. So $31^{12} < 2^{60} < 17^{15}$, and $17^{15}$ is greater.

Question 21 [Tier 2]

If $3^a = 4^b = 12^c$, find $c$ in terms of $a$ and $b$.

Answer. Let each one be $k$. Then $3 = k^{\frac{1}{a}}$, $4 = k^{\frac{1}{b}}$ and $12 = k^{\frac{1}{c}}$. Since 12 = 3 × 4, $k^{\frac{1}{c}} = k^{\frac{1}{a} + \frac{1}{b}}$, so $\frac{1}{c} = \frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}$. That gives $c = \frac{ab}{a + b}$.