Question

Giải các phương trình sau:

a) \({2^x} = \frac{1}{{{2^{x + 1}}}};\)                                

b) \(2{e^{2x}} = 5.\)

Answer 1

\(a,2^{3x-1}=2^{-\left(x+1\right)}\Rightarrow3x-1=-\left(x+1\right)\Rightarrow x=\dfrac{1}{2}\)

\(b,ln\left(2e^{2x}\right)=ln5\)

\(\Rightarrow ln2+lne^{2x}=ln5\)

\(\Rightarrow ln2+2x=ln5\)

\(\Rightarrow2x=ln5-ln2=ln\dfrac{5}{2}\)

Như vậy \(x=\dfrac{1}{2}ln\dfrac{5}{2}\)