Is this the equation?
\((\sqrt{7^}}) ^6x} = 49 ^{x-6}\)
First is to transform the two sides of the equation to have like bases.
\(7^{ \frac{6x}{2} } = 7^{2(x-6)}\)
Simplifying would result to:
\(7^{3x}} = 7^{2x-12}\)
Equating the exponents:
3x =2x+12
x= 12
If you would want to solve for x then x=12.
