Indices II
IB Mathematics AA · SL & HL · Shadow Worksheet
Practice
Question 1
Simplify, giving answers with positive indices:
(a)\(\dfrac{x^4 y^{-2}}{x^{-1} y^3}\)
(b)\(\left(\dfrac{2x^{-1}}{y^2}
ight)^{-3}\)
(c)\(\sqrt{x^6 y^4}\)
(a)
Answer
\(\dfrac{x^5}{y^5}\)
(b)
Apply \(-3\) power
\(\dfrac{y^6}{8x^{-3}} = \dfrac{x^3 y^6}{8}\)
Answer
\(\dfrac{x^3 y^6}{8}\)
Question 2
Solve the following equations:
(a)\(9^x = 3^{x+4}\)
(b)\(4^{2x-1} = 8^{x+1}\)
(c)\(25^x \cdot 5^{3-x} = 125\)
(a)
Base 3
\(3^{2x}=3^{x+4}\Rightarrow 2x=x+4\)
(b)
Base 2
\(2^{2(2x-1)}=2^{3(x+1)}\Rightarrow 4x-2=3x+3\)
(c)
Base 5
\(5^{2x}\cdot5^{3-x}=5^3\Rightarrow x+3=3\)
Question 3
Simplify \(\dfrac{(x^2 y)^3 \cdot x^{-4} y^{-1}}{(xy)^2}\).
Expand numerator
\(x^6 y^3 \cdot x^{-4}y^{-1} = x^2 y^2\); denominator \(x^2 y^2\)
Question 4
Evaluate without a calculator:
(a)\(27^{2/3}\)
(b)\(16^{-3/4}\)
(c)\(8^{1/3} \cdot 4^{3/2}\)
(a)
Cube root first
\((\sqrt[3]{27})^2 = 3^2\)
(b)
Fourth root first
\(\dfrac{1}{(\sqrt[4]{16})^3} = \dfrac{1}{2^3}\)
(c)
Simplify each
\(2 imes 8 = 16\)
Question 5
Solve \(x^{2/3} = 4\). Hence solve \(x^{2/3} - 5x^{1/3} + 4 = 0\) using the substitution \(u = x^{1/3}\).
Part 1
\(x^{2/3}=4\Rightarrow x=(4)^{3/2}=8\)
Substitution
\(u^2-5u+4=0\Rightarrow(u-1)(u-4)=0\Rightarrow u=1\) or \(u=4\)
Back-substitute
\(x^{1/3}=1\Rightarrow x=1\); \(x^{1/3}=4\Rightarrow x=64\)
Answer
\(x=1\) or \(x=64\)
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