Logarithm Basic (2)

$9^{2x-4}=27^{x}$ can be written as $3^{2(2x-4)}=3^{3x}$ now taking log both sides with base 3 $\log_{3} {3^{2(2x-4)}}=\log_{3} {3^{3x}}$ which simply gives $4x-8=3x$ hence $x=\boxed{8}$

$9^{2x}\times 9^{-4}=27^{x}\\3^{4x}\times 3^{-8}=3^{3x}\\3^{4x-8}=3^{3x}$ Equating $4x-8=3x\implies x=\boxed{8}$

(This solution uses logarithm because the topic says so) $\begin{array}{rcl} 9^{2x-4}&=&27^x\\ \log_39^{2x-4}&=&\log_327^x\\ (2x-4)\log_39&=& x\log_327\\ 2(2x-4)&=&3x\\ 4x-8&=&3x\\ x-8&=&0\\ x&=&8 \end{array}$

It took me a while to figure out what the question was lol ;)

3^(4x - 8) = 3^3x 4x - 8 = 3x x = 8

Just use the rules of logs.

Equating, you get 4x - 8 = 3x
thus, x = 8

\­( { 27 }^{ x }\quad =\quad { 9 }^{ x }\quad \times \quad { 3 }^{ x }\quad =\quad { 9 }^{ x }\quad \times \quad { 9 }^{ 1/2\quad x }\quad =\quad { 9 }^{ 3/2\quad x }\ \gg \quad 2x\quad -\quad 4\quad =\quad 3/2\quad x\quad \ \gg \quad x\quad =\quad 8 \­)