Question

Let $L$ denote ${\text{antilog}}_{32}0.6$ and $M$ denote the number of positive integers which have the characteristic $4$, when the base of $\log$ is $5$ and $N$ denote the value of ${49}^{(1-{\log }_{7}2)}+5^{-{\log }_{5}4}$. Find the value of $\frac{LM}{N}$.

Answer 1

$L={\text{antilog}}_{32}0.6={\left(32\right)}^{\frac{6}{10}}=2^{\frac{5\times 6}{10}}=2^3=8$
$M=$Integers from $625$ to $3125=2500$
$N={49}^{(1-{\log }_{7}2)}+5^{-{\log }_{5}4}$
$=49\times 7^{-2{\log }_{7}2}+5^{-{\log }_{5}4}$
$=49\times \frac{1}{4}+\frac{1}{4}=\frac{50}{4}=\frac{25}{2}[\because m\log a=\log a^m\text{\&}{a}^{{\log }_{a}b}=b]$
$\therefore$ $\frac{LM}{N}=\frac{8\times 2500\times 2}{25}=1600$
Ans: 1600