retagged by
444 views

1 Answer

1 votes
1 votes
Given that, $\log_{2}\left[3+\log_{3}\{4+\log_{4}(x-1)\}\right]-2=0$

$\Rightarrow \log_{2}\left[3+\log_{3}\{4+\log_{4}(x-1)\}\right]=2$

$\Rightarrow 3+\log_{3}\{4+\log_{4}(x-1)\}=2^{2}\quad \left[\because \log_{a}{b} = x \Rightarrow b=a^{x}\right]$

$\Rightarrow \log_{3}\{4+\log_{4}(x-1)\}=1$

$\Rightarrow 4+\log_{4}(x-1)=3^{1}$

$\Rightarrow \log_{4}(x-1)=-1$

$\Rightarrow x-1=4^{-1}$

$\Rightarrow x= \frac{1}{4}+1$

$\Rightarrow \boxed{x= \frac{5}{4}}$

$\therefore$ The value of $4x = 4 \times \frac{5}{4} = 5.$

Correct Answer $:5$
edited by
Answer:

Related questions

1 votes
1 votes
1 answer
1
soujanyareddy13 asked Jan 20, 2022
1,240 views
The number of ways of distributing $15$ identical balloons, $6$ identical pencils and $3$ identical erasers among $3$ children, such that each child gets at least four ba...
1 votes
1 votes
1 answer
2
1 votes
1 votes
1 answer
4
soujanyareddy13 asked Jan 20, 2022
883 views
For all possible integers $n$ satisfying $2.25 \leq 2 + 2^{n+2} \leq 202,$ the number of integer values of $3 + 3^{n+1}$ is