Question

Find the value of 'x' if $3\log 2+\frac{1}{3}\log 27-\log 4=\log x$

Answer 1

Given that:

$3\log 2+\frac{1}{3}\log 27-\log 4=\log x$

$[\because \log m^n=n\log m]$

Therefore

$\Rightarrow \log 2^3+\log {27}^{\frac{1}{3}}-\log 4=\log x$

$\Rightarrow \log 8+\log 3-\log 4=\log x$

$[\because \log m+\log n=\log (m\times n)]$

Therefore

$\Rightarrow \log (8\times 3)-\log 4=\log x$

$\Rightarrow \log 24-\log 4=\log x$

$[\because \log m-\log n=\log \left(\frac{m}{n}\right)]$

Therefore

$\Rightarrow \log \left(\frac{24}{4}\right)=\log x$

$\Rightarrow \log 6=\log x$

Now we take anti logarithm both side

$x=6$