I have this function: $$f(x)=\log(e^{2x}-5e^x+7)$$ I am supposed to find $f(x)=0$. I know that $log(x)=0$ when $x=1$. So I've put it this way: $$e^{2x}-5e^x+6=0$$ I don't know how to solve this, so I've tried substituting $e^x=t$ and I got $$t^2-5t+6=0$$ out of which I get two solutions: $t_1=3$ and $t_2=2$. When reverting back to $e^x$, I get: $e^x=3$ and $e^x=2$. Trying to solve for $x$, I get: $x_1=\log(3)$ and $x_2=\log(2)$. However, according to Wolfram Alpha this is not correct.
I don't know how to solve this, and I am not allowed calculator on the test.
Do you have any ideas? Thanks in advance!
Set $f(x) = 0$. Then,
$$0 = \log(e^{2x} - 5e^{x} + 7)$$
Set both sides by $e$ to get
$$\begin{aligned} e^{0} &= e^{\log(e^{2x} - 5e^{x} + 7)}\\ 1 &= e^{2x} - 5e^{x} + 7 \end{aligned}$$
So...
$$\begin{aligned} 0 &= e^{2x} - 5e^{x} + 7 - 1\\ 0 &= (e^x)^2 - 5e^x + 6\\ 0 &= (e^x - 2)(e^x - 3)\\ \end{aligned}$$
Thus,
$$\begin{aligned} e^x - 2 &= 0 & e^x - 3 &= 0\\ e^x &= 2 & e^x &= 3\\ x &= \ln(2) & x &= \ln(3) \end{aligned}$$
The points of intersection are $(\ln(2), 0)$ and $(\ln(3), 0)$