Question

Let $x=5^{\left({\log }_{5}2+{\log }_{5}3\right)}$.
If ' $d$ ' denotes the number of digits before decimal in $x^{30}$ and ' $c$ ' denotes the number of naughts after decimal before a significant digit starts in $x^{-20}$, then find the value of $(d-c)$.
[Given : ${\log }_{10}2=0.3010$ and ${\log }_{10}3=0.4771$.]

Answer 1

Answer: 0009

$x=2\cdot 3=6$
$\text{ Now, let }m=6^{30}$
${\log }_{10}m=30{\log }_{10}6=30(0.3010+0.4771)=0.7781\times 30=23.3430$
$\therefore d=24$
$\text{ Again let, }=6^{-20}$
$\log n=-20(0.7781)=-15.5620=\overline{16}.4379$
$\therefore c=15$
$\therefore (d-c)=24-15=9$