0 votes

$\log_{12}81=p$, then $3\bigg (\frac{4-p}{4+p}\bigg)$ is equal to

- $\log_416$
- $\log_68$
- $\log_616$
- $\log_28$

0 votes

Let $p= \log_{12} 81\implies \log_{12}3^4$

$\implies p=4 log_{12}3$

$\implies log_{12}3=\frac{p}{4} \qquad\dots\dots(i)$

Now $3 \bigg(\frac{4-p}{4+p} \bigg) = 3 \bigg(\frac{1-\frac{p}{4}}{1+\frac{p}{4}} \bigg)$

$\implies 3 \bigg(\frac{1-\log_{12}3}{1+log_{12}3}\bigg)$

$\implies 3 \bigg(\frac{\log_{12}12-\log_{12}3}{log_{12}12+log_{12}3}\bigg)$

$\implies 3 \bigg(\frac{\log (\frac{12}{3})}{\log (12*3)}\bigg)$

$\implies 3 \frac{\log 4}{\log 36}$

$\implies 3\log_{36}4=log_68$

Option $(B)$ is correct.

$\implies p=4 log_{12}3$

$\implies log_{12}3=\frac{p}{4} \qquad\dots\dots(i)$

Now $3 \bigg(\frac{4-p}{4+p} \bigg) = 3 \bigg(\frac{1-\frac{p}{4}}{1+\frac{p}{4}} \bigg)$

$\implies 3 \bigg(\frac{1-\log_{12}3}{1+log_{12}3}\bigg)$

$\implies 3 \bigg(\frac{\log_{12}12-\log_{12}3}{log_{12}12+log_{12}3}\bigg)$

$\implies 3 \bigg(\frac{\log (\frac{12}{3})}{\log (12*3)}\bigg)$

$\implies 3 \frac{\log 4}{\log 36}$

$\implies 3\log_{36}4=log_68$

Option $(B)$ is correct.