Logarithms and exponents with variables

Solution 1:

$log_a(b) = c$

This statement means $a^c = b$

This problem can be understood easily if you understand logarithms.

1. $log_x(a) = 0$ implies that $a = 1$ because $x^0 = 1$
2. $log_x(a) = 1$ implies that $a = x$ because $x^1 = x = a$

These two implications are used to solve the problem.

$log_x(log_{3x}(log_{6x}(2^{12}3^{12}4^68^49^627^4))) = 0$

By simplification, we get

$log_x(log_{3x}(log_{6x}(6^{36}))) = 0$

Using 1, we get

$log_{3x}(log_{6x}(6^{36})) = 1$

Using 2, we get

$log_{6x}(6^{36}) = 3x$

$(6x)^{3x} = 6^{36}$

By solving, we get $x = 6$

Solution 2:

The first few steps are the basic properties of logs. In the end, I think it's easier to go this way: $6^{36}=(6x)^{3x} \implies 6^{12}=(6x)^{x} \implies 1< x=6^y \le 12 \implies y=1, x=6$