Hi Mukulu.

Log₇ is how I write "log in base 7". And Log₁₄ is how I write "log in base 14".

So you have Log₇ (4x) = Log₁₄ (x).

Step 1: Let y = Log₁₄(x). This gives us:
Log₇ (4x) = y.

Step 2: Raise both sides under the base of 7:
7^{[Log₇ (4x)]} = 7^y
4x = 7^y

Step 3: Since y = Log₁₄(x), let's raise this equation under base 14:
14^y = 14^{[Log₁₄(x)]}
14^y = x.

Step 4: Substitute x from Step 3 into Step 2 equations:
4 [ 14^y] = 7^y
4 (2^y)(7^y) = 7^y
4 (2^y) = 1

Can you finish the problem now? x works out to a nice simple number.

Cheers,
Stephen La Rocque.