Question

\(2^{x^2-2x}.3^x=\frac{3}{2}\)

Answer 1

ta có

\(2^{x^2-2x}.3^x=\frac{3}{2}\Leftrightarrow2^{x^2-2x}.2=3=3^x\Leftrightarrow2^{x^2-2x+1}=3^{1-x}\Leftrightarrow2^{\left(x-1\right)^2}=3^{1-x}\)

lấy logarit cơ số 2 của 2 vế ta đc

\(\left(x-1\right)^2=\left(1-x\right)\log_23\Rightarrow\left(1-x\right)^2-\left(1-x\right)\log_23=0\Leftrightarrow\left(1-x\right)\left(1-x-\log_23\right)=0\)

suy ra 1-x=0 suy ra x=1

hoặc \(1-x-\log_23=0\Leftrightarrow x=1-\log_23\)