GATE Overflow Test Series | Quantitative Aptitude | Test 1 | Question: 18

Question

If $x = \sqrt{7\sqrt{7\sqrt{7\dots \infty}}},$ then the value of $\log_{49^{3}} x$ is _________

• A. $\dfrac{1}{7}$
• B. $\dfrac{1}{6}$
• C. $\dfrac{1}{3}$
• D. $6$

Answer 1

Given that $x =\sqrt{7\sqrt{7\sqrt{7\dots \infty}}}$

$\implies x = \sqrt{7x}$

$\implies x^{2} = 7x$

$\implies x^{2} - 7x = 0$

$\implies x(x - 7) = 0$

$\implies x \neq 0\,(\text{because it is not satisfied  the given equation})$

$\implies x = 7$

Now, $\log_{49^{3}} x = \log_{(7^{2})^{3}} 7 = \log_{7^{6}} 7 $

$= \dfrac{1}{6} \qquad(\because \left({7^6}\right)^{1/6} = 7)$

Last step can also be solved by

$\log_{7^{6}} 7= \dfrac{1}{6} \log_{7} 7 = \dfrac{1}{6}.\qquad [\because\log_{a^{b}} c = \dfrac{1}{b}\log_{a} c,\,\, \log_{a} a = 1]$

So, the correct answer is $(B).$