Indices (AQA GCSE Further Maths): Revision Notes

Worked Example: Express as single powers of x

For $\frac{x^3 \times x^2}{x^7}$:

• First, use Law 1 to combine the numerator: $x^3 \times x^2 = x^{3+2} = x^5$
• Then use Law 2 to divide: $x^5 \div x^7 = x^{5-7} = x^{-2}$

For $\sqrt\[3]{x^5 \times x^3}$:

• Combine the terms inside: $x^5 \times x^3 = x^8$
• Express the cube root as a fractional index: $\sqrt\[3]{x^8} = (x^8)^{1/3}$
• Apply Law 3: $(x^8)^{1/3} = x^{8 \times 1/3} = x^{8/3}$

Worked Example: Solve $x^{2/3} = 8$

Method 1 (Using roots and powers):

• Square both sides: $(x^{2/3})^2 = 8^2$, giving $x^{4/3} = 64$
• Take the cube root: $x = \sqrt\[3]{64} = 4$

Method 2 (Using fractional powers):

• Raise both sides to the power of $3/2$: $(x^{2/3})^{3/2} = 8^{3/2}$
• This gives $x^1 = 64^{1/2} = \sqrt{64} = 4$

Worked Example: Solve $x^{-1/3} = 10$

Method 1 (Using reciprocals):

• Take the reciprocal of both sides: $x^{1/3} = \frac{1}{10}$
• Cube both sides: $x = \left(\frac{1}{10}\right)^3 = \frac{1}{1000}$

Worked Example: Solving a disguised quadratic

Since $x^4 = (x^2)^2$, we can use the substitution $y = x^2$ to transform the equation:

• Original equation: $x^4 - 5x^2 + 4 = 0$
• After substitution: $y^2 - 5y + 4 = 0$
• Factoring: $(y - 4)(y - 1) = 0$
• So $y = 4$ or $y = 1$
• Since $y = x^2$, we have $x^2 = 4$ or $x^2 = 1$
• Therefore $x = \pm 2$ or $x = \pm 1$

Worked Example: Solve $4^x = 2^x + 56$

This equation demonstrates another type of disguised quadratic. The key insight is recognising that $4^x$ can be rewritten using the same base as $2^x$.

Since $4^x = (2^2)^x = 2^{2x} = (2^x)^2$, we can substitute $y = 2^x$:

• The equation becomes: $y^2 = y + 56$
• Rearranging: $y^2 - y - 56 = 0$
• Factoring: $(y - 8)(y + 7) = 0$
• So $y = 8$ or $y = -7$

However, since $y = 2^x$ and exponential functions are always positive, $y = -7$ is impossible. Therefore $y = 8$, which means $2^x = 8 = 2^3$, so $x = 3$.