If x is a real number such that \sqrt{\frac{25^{x+1} - 25^x}{5^{x+2} + 5^{x+1}}} = 250 wha… — GRE General Test

x

is a real number such that

\sqrt{\frac{25^{x+1} - 25^x}{5^{x+2} + 5^{x+1}}} = 250

what is the value of

x

Cevap: 7

The value of

x

is 7.

Factoring out common powers in the numerator and denominator yields

25^x(25-1) = 24 \cdot 5^{2x}

and

5^{x+1}(5+1) = 6 \cdot 5^{x+1}

. Simplifying their ratio inside the square root gives

\frac{24 \cdot 5^{2x}}{6 \cdot 5^{x+1}} = 4 \cdot 5^{x-1}

. Taking the square root gives

2 \cdot 5^{(x-1)/2}

. Setting this equal to 250 yields

5^{(x-1)/2} = 125 = 5^3

. Equating the exponents gives

(x-1)/2 = 3

, which solves to

x = 7

Factor the numerator and express terms with a common base of 5

25^{x+1} - 25^x = 25^x(25 - 1) = 24 \cdot (5^2)^x = 24 \cdot 5^{2x}

Factoring out

25^x

simplifies the difference into a single term with base 5.

Factor the denominator using base 5

5^{x+2} + 5^{x+1} = 5^{x+1}(5 + 1) = 6 \cdot 5^{x+1}

Factoring out the common power

5^{x+1}

simplifies the sum into a single term.

Simplify the fraction inside the square root

\frac{24 \cdot 5^{2x}}{6 \cdot 5^{x+1}} = 4 \cdot 5^{2x - (x+1)} = 4 \cdot 5^{x-1}

Dividing coefficients (24/6 = 4) and applying exponent rules for division (

a^m / a^n = a^{m-n}

Take the square root of the simplified expression

\sqrt{4 \cdot 5^{x-1}} = \sqrt{4} \cdot \sqrt{5^{x-1}} = 2 \cdot 5^{\frac{x-1}{2}}

Using radical rules

\sqrt{ab} = \sqrt{a}\sqrt{b}

and

\sqrt{a^k} = a^{k/2}

Set the simplified radical expression equal to 250 and solve for x

2 \cdot 5^{\frac{x-1}{2}} = 250 \implies 5^{\frac{x-1}{2}} = 125 \implies 5^{\frac{x-1}{2}} = 5^3 \implies \frac{x-1}{2} = 3 \implies x = 7

Dividing both sides by 2 gives

5^{(x-1)/2} = 125 = 5^3

. Equating exponents yields

(x-1)/2 = 3

, so

x = 7

Exponent rules, base conversion, factoring exponential terms, and radical simplification

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