GRE (3)

Calculus Level 2

e a e d x x a a x d y y = 1 {\Large \int_{e}^{a^e}} \frac{dx}{\displaystyle x \int_{a}^{ax}\frac{dy}{y}} =1

What positive value of a \color{#D61F06} a satisfies the equation above?

e 2 e^2 e 1 e e^{\frac{1}{e}} 1 e \frac{1}{e} e \sqrt{e} e e

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2 solutions

Hana Wehbi
May 5, 2018

e a e d x x a a x d y y = 1 \large \int_{e}^{a^e} \frac{dx}{x \int_{a}^{ax}\frac{dy}{y}} =1

e a e d x x ln y a a x = 1 \large \int_{e}^{a^e} \frac{dx}{x \ln y|_a^{ax }}=1

e a e d x x ln x = 1 \large \int_{e}^{a^e} \frac{dx}{x \ln x} =1

ln ( ln x ) e a e = 1 \large \ln( \ln x)|_e^{a^e}=1

ln ( ln a e ) = 1 \large \implies \large \ln (\ln{a^e})= 1

ln a e = e a = e \large\implies \ln a^e= e \implies \boxed {\large a=e }

Similar solution with @Hana Nakkache 's

I = e a e d x x a a x d y y = e a e d x x ln y a a x = e a e d x x ( ln a x ln a ) = e a e d x x ln x Let u = ln x d u = d x x = 1 e ln a d u u = ln u 1 e ln a = ln a \begin{aligned} I & = \int_e^{a^e} \frac {dx}{x\int_a^{ax} \frac {dy}y} = \int_e^{a^e} \frac {dx}{x\ln y\ \big|_a^{ax}} \\ & = \int_e^{a^e} \frac {dx}{x(\ln ax - \ln a)} = \int_e^{a^e} \frac {dx}{x\ln x} & \small \color{#3D99F6} \text{Let }u = \ln x \implies du = \frac {dx}x \\ & = \int_1^{e\ln a} \frac {du}u = \ln u\ \bigg|_1^{e\ln a} = \ln a \end{aligned}

Therefore, I = ln a = 1 I=\ln a = 1 a = e \implies a = \boxed{e} .

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