Question

If $3^{4x+1}={\left(27\right)}^{x+1}$, then find $x$.

Answer 1

$3^{4x+1}={\left(27\right)}^{x+1}$

$\Rightarrow 3^{4x+1}={\left(3^3\right)}^{x+1}$

$\Rightarrow 3^{4x+1}=3^{3x+3}$ $(\because {\left(a^m\right)}^n=a^{mn})$

$\Rightarrow \frac{3^{4x+1}}{3^{3x+3}}=1$

$\Rightarrow 3^{4x+1-3x-3}=1$ $(\because \frac{a^m}{a^n}=a^{m-n})$

$\Rightarrow 3^{x-2}=3^0$ $(\because a^0=1)$

$\Rightarrow x-2=0$

$\Rightarrow x=2$

Hence, the answer is $2.$