Question

If ${\log }_{10}5=a$ and ${\log }_{10}3=b$, then which of the following is/are correct?

• A. ${\log }_{30}8=\frac{3(1-a)}{b+1}$
• B. ${\log }_{40}15=\frac{a+b}{3-2a}$
• C. ${\log }_{243}32=\frac{1-a}{b}$
• D. none of these

Answer 1

Answer: (A), (B), (C)

${\log }_{40}15=\frac{a+b}{3-2a}$
${\log }_{30}8=\frac{3(1-a)}{b+1}$
${\log }_{243}32=\frac{1-a}{b}$

${\log }_{10}5=a$
${\log }_{10}3=b$
let $x=\frac{3(1-a)}{b+1}$
$\Rightarrow x=\frac{3({\log }_{10}10-{\log }_{10}5)}{{\log }_{10}3+{\log }_{10}10}=\frac{3{\log }_{10}\left(\frac{10}{5}\right)}{{\log }_{10}(3\times 10)}$
$\Rightarrow x=\frac{3{\log }_{10}\left(2\right)}{{\log }_{10}\left(30\right)}={\log }_{30}8$
let $y=\frac{a+b}{3-2a}$
$\Rightarrow y=\frac{{\log }_{10}5+{\log }_{10}3}{3{\log }_{10}10-2{\log }_{10}5}=\frac{{\log }_{10}(5\times 3)}{{\log }_{10}\left(\frac{{10}^3}{5^2}\right)}$
$\Rightarrow y={\log }_{40}15$
let $z=\frac{1-a}{b}$
$\Rightarrow z=\frac{{\log }_{10}10-{\log }_{10}5}{{\log }_{10}3}=\frac{{\log }_{10}\left(\frac{10}{5}\right)}{{\log }_{10}3}$
$\Rightarrow z=\frac{5{\log }_{10}2}{5{\log }_{10}3}=\frac{{\log }_{10}32}{{\log }_{10}243}$
$\Rightarrow z={\log }_{243}32$
Ans: A,B,C