If x, y, and z are non-zero real numbers satisfying the exponential equation 2^x = 5^y = 1… — JAMB UTME

If $x$ , $y$ , and $z$ are non-zero real numbers satisfying the exponential equation $2^x = 5^y = 100^z$ , what is the numerical value of the expression $z \left( \frac{2}{x} + \frac{2}{y} \right)$ ?

Answer: 1

Answer

The numerical value of

z \left( \frac{2}{x} + \frac{2}{y} \right)

1

By setting

2^x = 5^y = 100^z = k

, we can write

2 = k^{1/x}

5 = k^{1/y}

, and

100 = k^{1/z}

. Factoring

100 = 2^2 \times 5^2

gives

k^{1/z} = (k^{1/x})^2 \times (k^{1/y})^2 = k^{2/x + 2/y}

. Equating exponents gives

\frac{1}{z} = \frac{2}{x} + \frac{2}{y}

, which upon multiplying by

z

yields

1

Step-by-Step Solution

Equate the given exponential expressions to a common constant

k

2^x = 5^y = 100^z = k

Introducing a common variable allows isolating each base exponent combination.

Express bases

2

5

, and

100

in terms of

k

using fractional indices.

2 = k^{\frac{1}{x}}

5 = k^{\frac{1}{y}}

100 = k^{\frac{1}{z}}

Applying the power law

(a^m)^{\frac{1}{m}} = a

isolates each base.

Express

100

using prime factorization of the other bases.

100 = 2^2 \times 5^2

Establishing a numerical relationship between

100

2

, and

5

links the exponential variables.

Substitute the

k

-expressions into

100 = 2^2 \times 5^2

and apply index multiplication laws.

k^{\frac{1}{z}} = \left(k^{\frac{1}{x}}\right)^2 \times \left(k^{\frac{1}{y}}\right)^2 = k^{\frac{2}{x}} \times k^{\frac{2}{y}} = k^{\frac{2}{x} + \frac{2}{y}}

Multiplying powers with the same base requires adding the exponents:

a^m \cdot a^n = a^{m+n}

Equate exponents of identical bases.

\frac{1}{z} = \frac{2}{x} + \frac{2}{y}

k^a = k^b

for

k > 1

, then

a = b

Multiply both sides of the equation by

z

z \left( \frac{2}{x} + \frac{2}{y} \right) = 1

Rearranging the equation yields the exact numerical value of the requested expression.

Key Concept

Equating Exponents of Common Bases and Fractional Indices

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Answer

Step-by-Step Solution

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