Find the value of x if ≤ft[ {1}{64^{x+2}} ] = ≤ft[ {4^{x-3}}{16^{x}} ]

Correct Answer is : −32-3/2−23

Correct Answer is : −32-3/2−23

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JAMB Mathematics 2015 Paper

Question : 29

Total: 61

\left[ \frac{1}{64^{x+2}} \right] = \left[ \frac{4^{x-3}}{16^{x}} \right]

Solution:

Concept:

Express both sides as powers of the same base so that the exponents can be compared directly.

Explanation:

Start with the equation:

\frac{1}{64^{x+2}} = \frac{4^{x-3}}{16^{x}}

Rewrite

64

4

, and

16

as powers of

2

64 = 2^6

4 = 2^2

16 = 2^4

The left-hand side becomes:

\frac{1}{(2^6)^{x+2}} = 2^{-6(x+2)} = 2^{-6x-12}

The right-hand side becomes:

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

Since the bases are equal, equate the exponents:

-6x - 12 = -2x - 6

Collect like terms:

-6x + 2x = -6 + 12

-4x = 6

Divide both sides by

-4

x = -\frac{6}{4} = -\frac{3}{2}

Answer:

x = -\frac{3}{2}

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