Question

If  \(log_2[3 + log_ 3[4 + log_4(x - 1)] - 2 = 0\) then 4x equals

Answer 1

Given: 
\(\log_2 \left[ 3 + \log_3 \left( 4 + \log_4 (x-1) \right) \right] - 2 = 0\)
Now, rearranging and simplifying: 
\(\log_2 \left[ 3 + \log_3 \left( 4 + \log_4 (x-1) \right) \right] = 2\)

Using the properties of logarithm: \(3 + \log_3 \left( 4 + \log_4 (x-1) \right) = 2^2\)
\(3 + \log_3 \left( 4 + \log_4 (x-1) \right) = 4\)

Subtracting 3 from both sides:
\(\log_3 \left( 4 + \log_4 (x-1) \right) = 1\)

This implies: \(4 + \log_4 (x-1) = 3\) 
\(\log_4 (x-1) = -1\)

Now, using the properties of logarithm:
\(x-1 = 4^{-1}\)

\(x-1 = \frac{1}{4}\)

Now, adding 1 to both sides: 
\(x = \frac{5}{4}\)

To find \(4x\): \(\ 4x = 4 \times \frac{5}{4} = 5\)

Answer 2

Given:
\(\log_2(3+\log_3(4+\log_4(x-1)))-2=0\)
Let's Simplify the equation:
\(\log_2(3+\log_3(4+\log_4(x-1)))=2\)
Now using the definition of logarithms, \(b^y=x, then\;y=\log_b(x)\)
\(3+\log_3(4+\log_4(x-1))=2^2\)
\(3+\log_3(4+\log_4(x-1))=4\)
\(\log_3(4+\log_4(x-1))=4-3\)
\(\)\(\log_3(4+\log_4(x-1))=1\)
Now use the definition of logarithms again:
\(4+\log_4(x-1)=3^1\)
\(\log_4(x-1)=3-4\)
\(\log_4(x-1)=-1\)
Again we use the definition of logarithms:
\(x-1=4^{-1}\)

\(x-1=\frac{1}{4}\)

\(x=\frac{1}{4}+1\)

\(x=\frac{1}{4}+\frac{4}{4}\)

\(x=\frac{5}{4}\)

\(4x=5\)
So, the answer is 5.