Question

Solve and find the value of $x$ : $4^{x-1}-3\cdot 2^{x-1}+2=0$

Answer 1

We have,

$4^{x-1}-3\cdot 2^{x-1}+2=0$

$2^{2(x-1)}-3\cdot 2^{x-1}+2=0$

Let $t=2^{x-1}$

Then,

$t^2-3t+2=0$

$t^2-2t-t+2=0$

$(t-1)(t-2)=0$

$t=1,2$

So,

$2^{x-1}=1$ or $2^{x-1}=2$

$\frac{2^x}{2}=1$ or $x-1=1$

$2^x=2$ or $x=2$

$x=1$ or $x=2$

Hence, this is the answer.