15.8=2(e^(x/4) + e^(-x/4)) Solve for X
Hello, OzzMan!
15) Solve for $\displaystyle x\!:\;\;2e^{\frac{x}{4}} + e^{-\frac{x}{4}} \;=\;8$
Multiply through by $\displaystyle e^{\frac{x}{4}}\!:\;\;2e^{\frac{x}{2}} + 1 \;=\;8e^{\frac{x}{4}}\quad\Rightarrow\quad2\left(e ^{\frac{x}{4}}\right)^2 - 8\left(e^{\frac{x}{4}}\right) + 1 \;=\;0$
Let $\displaystyle u = e^{\frac{x}{4}}$ and we have: .$\displaystyle 2u^2 - 8u + 1 \:=\:0$
Quadratic Formula: .$\displaystyle u \;=\;\frac{4\pm\sqrt{14}}{2}$
Back-substitute: .$\displaystyle e^{\frac{x}{4}} \;=\;\frac{4\pm\sqrt{14}}{2} \quad\Rightarrow\quad \frac{x}{4} \;=\;\ln\left(\frac{4 \pm\sqrt{14}}{2}\right)$
Therefore: .$\displaystyle x \;=\;4\!\cdot\!\ln\left(\frac{4\pm\sqrt{14}}{2}\ri ght) $
my bad not 15.8. its 5.8. doesnt really matter much there though. but my 2 answers from quadratic formula were 2.5 and .4
when i plugged them into y=(x/4) i didnt get the same answer as when i plugged it into 4(ln(x))=y
arent these the same ? the right answer comes out of the 2nd one i listed.
The original equation is:
$\displaystyle 5.8=2\left(e^{\frac x4} + e^{-\frac x4}\right)$
With the substitution $\displaystyle y = e^{\frac x4}$ the equation becomes
$\displaystyle 5.8=2\left(y + \frac1y\right)$ ........which you solved correctly:
y = 2.5 or y = 0.4
If you plug in these values you'll get:
$\displaystyle 5.8=2\left(2.5 + \frac1{2.5}\right)=2 \cdot 2.5 + 2 \cdot 0.8 = 5.8$
$\displaystyle 5.8=2\left(0.4 + \frac1{0.4}\right)=2 \cdot 0.8 + 2 \cdot 2.5 = 5.8$
Ahemmm....
Yeah, so y=2.5 or y=.4, but you have set y=e^(x/4)
which means that ln(y)=x/4
which is that x=4ln(y) then you now plug in your values for y to get x which is what you are actually solving for. Which then matches the book answers.
(Sorry for not taking the time to actually to actually figure out the latex generator or whatever is used here.)