Find the value of x if \( [1 \div 64^{(x+2)}] = [4^{(x-3)} \div 16^x] \)

Assessment: JAMB UTME - Mathematics - 2015 Subject: General Mathematics

Question 1 Report

\frac{3}{2}

\frac{2}{3}

\frac{1}{3}

- \frac{3}{2}

Answer Details

[1 \div 64^{(x + 2)}] = [4^{(x - 3)} \div 16^{x}]

64^{- (x + 2)} = [4^{(x - 3)}] \div [16^{x}]

Break down 4, 16 and 64 into smaller index numbers:

2^{- 6 (x + 2)} = 2^{2 (x - 3)} \div 2^{4 (x)}

2^{- 6 x - 12} = 2^{2 x - 4 x - 6}

2^{- 6 x - 12} = 2^{- 2 x - 6}

− 6x − 12 = − 2x − 6
Collect the like terms:
−6x + 2x = −6 + 12
−4x =6
x =

\frac{6}{4}

x =

\frac{- 3}{2}

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