retagged by
545 views

1 Answer

0 votes
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.

Related questions

2 votes
2 votes
1 answer
1
go_editor asked Mar 19, 2020
575 views
If $\log_2(5+\log_3a)=3$ and $\log_5(4a+12+\log_2b)=3$, then $a+b$ is equal to$67$$40$$32$$59$
2 votes
2 votes
1 answer
2
go_editor asked Mar 19, 2020
553 views
If $x$ is a positive quantity such that $2^x=3^{\log_52}$, then $x$ is equal to$1+\log_3\dfrac{5}{3}$$\log_58$$1+\log_5\dfrac{3}{5}$$\log_59$
2 votes
2 votes
1 answer
3
go_editor asked Mar 19, 2020
560 views
Let $x, y, z$ be three positive real numbers in a geometric progression such that $x < y < z$. If $5x$, $16y$, and $12z$ are in an arithmetic progression then the common ...
3 votes
3 votes
1 answer
5
go_editor asked Mar 19, 2020
731 views
Given that $x^{2018}y^{2017}=1/2$ and $x^{2016}y^{2019}=8$, the value of $x^2+y^3$ is$35/4$$37/4$$31/4$$33/4$