How do you calculate this?

Question

$$e^x+3e^{-x}=4 $$please can you show me how to ln this or something?

Answer 1

consider $e^x=y$, then our equation becomes $y+\frac 3y=4$

$y^2-4y+3=0\\ y^2-3y-y+3=0\\ y(y-3)-(y-3)=0\\ (y-1)(y-3)=0$

Therefore, either $y=1 \text{ or }y=3\\ e^x=1 \text{ or } e^x=3\\ x=ln1 \text{ or }x=ln3$

Hence, the solution is $x=0 \text{ or }x=ln3$

I hope this helped...

Answer 2

Hint:

As $e^x\ne0,$

$$(e^x)^2-4(e^x)+3=0$$

$$e^x=?$$

Answer 3

Write $e^x=t$ and multiply by $t$; see how the equation changes into a typical algebraic equation.

Answer 4

Set $y=e^x$ then $y+1/y=4$....solve for $y$ and then find $x$.

Answer 5

The equation is equivalent to $e^{2x} + 3 = 4e^x$. Let $e^x = y$. Then, $y^2 - 4y + 3 = 0$. On solving for $y$, we get y equals either 3 or 1. Which means $e^x$ equals 3 or 1. Applying natural log, we get solution of $x$ as either $ln 3$ or $ln 1$ (which is 0).