Question

# Solve: e^2x=3(e^x)+4

Equations and inequalities
Solve:
$$\displaystyle{e}^{{2}}{x}={3}{\left({e}^{{x}}\right)}+{4}$$

2021-08-16
We have to solve $$\displaystyle{e}^{{2}}{x}={3}{\left({e}^{{x}}\right)}+{4}$$ Let $$\displaystyle{u}={e}^{{x}}$$
Then given equation come:
$$\displaystyle{u}^{{2}}={3}{u}+{4}\Rightarrow{u}^{{2}}-{3}{u}-{4}={0}$$
By using quadratic formula we get
$$\displaystyle{u}={\frac{{-{\left(-{3}\right)}±\sqrt{{-{3}}}^{{2}}-{4}\cdot{1}\cdot{\left(-{4}\right)}}}{{{2}\cdot{1}}}}={\frac{{{3}±\sqrt{{{9}+{16}}}}}{{{2}}}}={4},{1}$$
Therefore we get
$$\displaystyle{e}^{{x}}={4}{\quad\text{and}\quad}{e}^{{x}}={1}$$
By taking logarithm in the above two equations we get
$$\displaystyle{x}={\ln{{\left({4}\right)}}}$$ and $$\displaystyle{x}={\ln{{\left({1}\right)}}}={0}$$
Notice that x=0 does not satisfy the given equation.
Hence the only solution of the given equation is x=ln(4)=2ln(2)