Logarithm Basic

$3^{3x}=9^{2x-1}$ can be written as $3^{3x}=3^{2(2x-1)}$ hence taking log with base 3 on both sides $\log_{3} {3^{3x}}=\log_{3} {3^{2(2x-1)}}$ which gives $3x= 4x-2$ hence $x=\boxed{2}$

$Given\quad \quad that\\ \qquad { 3 }^{ 3x }={ 9 }^{ 2x-1 }\\ or\quad { 3 }^{ 3x }={ ({ 3 }^{ 2 }) }^{ (2x-1) }\\ or\quad { 3 }^{ 3x }={ 3 }^{ 4x-2 }\\ As\quad Bases\quad are\quad same\quad on\quad both\quad sides\\ So,\quad Powers\quad should\quad be\quad same.\\ =>\quad 3x=4x-2\\ or\quad \quad 2=4x-3x\\ or\quad \quad \boxed { x=2 }$

The 9^(2x-1) part could be expressed with base 3 by multiplying the exponents by 2. Therefore 3x=2(2x-1) and x=2.

9^(2x-1)= 3^2(2x-1)=3^3x

i.e. 4x-2=3x

So x=2