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JAMB Mathematics 2019 Paper

Question : 50

Total: 73

25^{1 - x} \times 5^{x + 2} \div \left(\frac{1}{125}\right)^{x} = 625^{-1}

, find the value of x.

Solution:

Concept:

Express every number as a power of

5

, then equate the exponents since the bases are equal.

Explanation:

Rewrite each base as a power of

5

25 = 5^2

\frac{1}{125} = 125^{-1} = (5^3)^{-1} = 5^{-3}

, and

625 = 5^4

So the equation becomes:

(5^2)^{1-x} \times 5^{x+2} \div (5^{-3})^{x} = (5^4)^{-1}

Apply the laws of indices:

(a^m)^n = a^{mn}

and

a^m \div a^n = a^{m-n}

5^{2(1-x)} \times 5^{x+2} \div 5^{-3x} = 5^{-4}

5^{2-2x} \times 5^{x+2} \div 5^{-3x} = 5^{-4}

Combine the exponents on the left-hand side:

5^{(2-2x) + (x+2) - (-3x)} = 5^{-4}

Since the bases are equal, equate the exponents:

2 - 2x + x + 2 + 3x = -4

Simplify:

4 + 2x = -4

2x = -8

x = -4

Answer:

x = -4

Option A is correct.

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