Question

The sum of digits of antilogarithm of $0.75$, if the base of the logarithm is $2401$ is

Answer 1

Let the number be $x$.

Given, anti-logarithm of $x$ to the base $2401$ is $0.75$.

$\Rightarrow x={2401}^{0.75}$

$\Rightarrow x=(2401{)}^{\frac{3}{4}}$

$\Rightarrow x={7^4}^{\left(\frac{3}{4}\right)}$

$\Rightarrow x=7^3=343$

Sum of the digits of $x=3+4+3=10$