Malliratkaisuja

423

f\left(x\right)=e^{2x}-25

f\left(x\right)=0
e^{2x}-25=0
e^{2x}=25
\left(e^x\right)^2=25
kolme tapaa jatkaa:
\begin{array}{l|l} e^x=5\ \ \ \ \text{tai}\ \ \ e^x=-5&\ln25=2x&\ln5^2=2x\\ \hline x=\ln5\ \ \text{tai}\ ei\ ratk.&x=\frac{\ln25}{2}=\frac{\ln5^2}{2}=\frac{2\ln5}{2}=\ln5&2\ln5=2x\\ &&\ln5=x \end{array}