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

$ 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 $
$\therefore(d-c)=24-15=9$