Question

log2(5+log3a)=3 log5(4a+12+log2b)=3 then a+b equal to​

Answer 1

$\textbf{Given:}$

$\log_2(5+\log_3a)=3$

$\log_5(4a+12+\log_2b)=3$

$\textbf{To find:}$

$\text{The value of a+b }$

$\textbf{Solution:}$

$\text{Consider,}$

$\log_2(5+\log_3a)=3$

$\text{Convert it into exponential form, we get}$

$5+\log_3a=2^3$

$\log_3a=8-5$

$\log_3a=3$

$a=3^3$

$\implies\boxed{\bf\,a=27}$

$\text{Consider,}$

$\log_5(4a+12+\log_2b)=3$

$\text{Convert it into exponential form, we get}$

$4a+12+\log_2b=5^3$

$4(27)+12+\log_2b=125$

$108+12+\log_2b=125$

$120+\log_2b=125$

$\log_2b=5$

$b=2^5$

$\implies\boxed{\bf\,b=32}$

$\text{Now,}$

$a+b=27+32=59$

$\therefore\textbf{The value of a+b is 59}$

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