# Math Help - Logarithm equation

1. ## Logarithm equation

15.8=2(e^(x/4) + e^(-x/4)) Solve for X

2. Originally Posted by OzzMan
15.8=2(e^(x/4) + e^(-x/4)) Solve for X
Use the substitution $y = e^{\frac x4}$

$15.8 = 2\left( y + \frac1y \right)$

Solve for y. Then re-substitute and solve for x.

3. Hello, OzzMan!

15) Solve for $x\!:\;\;2e^{\frac{x}{4}} + e^{-\frac{x}{4}} \;=\;8$

Multiply through by $e^{\frac{x}{4}}\!:\;\;2e^{\frac{x}{2}} + 1 \;=\;8e^{\frac{x}{4}}\quad\Rightarrow\quad2\left(e ^{\frac{x}{4}}\right)^2 - 8\left(e^{\frac{x}{4}}\right) + 1 \;=\;0$

Let $u = e^{\frac{x}{4}}$ and we have: . $2u^2 - 8u + 1 \:=\:0$

Quadratic Formula: . $u \;=\;\frac{4\pm\sqrt{14}}{2}$

Back-substitute: . $e^{\frac{x}{4}} \;=\;\frac{4\pm\sqrt{14}}{2} \quad\Rightarrow\quad \frac{x}{4} \;=\;\ln\left(\frac{4 \pm\sqrt{14}}{2}\right)$

Therefore: . $x \;=\;4\!\cdot\!\ln\left(\frac{4\pm\sqrt{14}}{2}\ri ght)$

4. 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.

5. Originally Posted by OzzMan
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.
Hi,

If you want to use the substitution you must use: $
y = e^{\frac x4}
$
which is not the same as $y = \frac x4$

7. wait. yeah i did use e^(x/4). i did e^(2.5/4) got 1.86. When i plug 2.5 into 4ln(2.5) i get 3.66 which is the correct answer. Is there anything im doing wrong?

8. Originally Posted by OzzMan
wait. yeah i did use e^(x/4). i did e^(2.5/4) got 1.86. When i plug 2.5 into 4ln(2.5) i get 3.66 which is the correct answer. Is there anything im doing wrong?
The original equation is:

$5.8=2\left(e^{\frac x4} + e^{-\frac x4}\right)$

With the substitution $y = e^{\frac x4}$ the equation becomes

$5.8=2\left(y + \frac1y\right)$ ........which you solved correctly:

y = 2.5 or y = 0.4

If you plug in these values you'll get:

$5.8=2\left(2.5 + \frac1{2.5}\right)=2 \cdot 2.5 + 2 \cdot 0.8 = 5.8$

$5.8=2\left(0.4 + \frac1{0.4}\right)=2 \cdot 0.8 + 2 \cdot 2.5 = 5.8$

Ahemmm....

9. the books answer though is plus or minus 3.66

10. 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.)

11. ah thanks to all.