Question

Prove that

$\left(i\right)log12=log3+log4$

$\left(ii\right)log50=log2+2log5$

$\left(iii\right)log(1+2+3)=log1+log2+log3$

Answer 1

(i) We have,

$12=3\times 4$

$\therefore log(3\times 4)$

$\Rightarrow log12=log3+log4[\because logmn=logm+logn]$

Thus, $log12=log3+log4$.

RHS= $log3+log4=log(3\times 4)=log12=LHS$

(ii) We have
$50=2\times 5^2$
$\therefore log50=log(2\times 5^2)$
$=log2+log{5}^2[\because logmn=logm+logn]=log2+2log5[\because log{m}^n=nlogm]$

ALTER RHS$=log2+2log5$
$=log2+log{5}^2[\because nlogm=log{m}^n]=log(2\times 5^2)[\because logm+logn=logmn]=log50=LHS$

(iii) We have,
LHS$=log(1+2+3)=log6=log(1\times 2\times 3)[\because 6=1\times 2\times 3]=log1+log2+log3\left[\begin{array}{c}\because log\left(mnp\right)\\ =logm+logn+logp\end{array}\right]$

=RHS
Hence, $log(1+2+3)=log1+log2+log3$