Question

Solve the equation:
e^2x - e^x - 20 = 0

Answer 1

substitute e^x = u
and then plug it in the equation and solve it

Answer 2

a quadratic equation

\[\rm ax^2+bx+c=0 \\roots~will~be~: x=\large\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

Answer 3

\(\rm Here~e^x=x\)

Answer 4

but dont you have to use logs to cancel out the e? (or ln)

Answer 5

first find out the root

Answer 6

and how would I do that?

Answer 7

You need to first apply your exponent rule: \(\large\rm e^{2x}=(e^x)^2\)
That should make it easier to see why a substitution is appropriate.

Answer 8

\[\large\rm e^{2x} - e^x - 20 = 0\]Becomes,\[\large\rm (e^{x})^2 - e^x - 20 = 0\]And then if you make a substitution,\[\large\rm (\color{orangered}{e^x})^2 - \color{orangered}{e^x} - 20 = 0\]it becomes a simple quadratic, ya?\[\large\rm (\color{orangered}{u})^2 - \color{orangered}{u} - 20 = 0\]

Answer 9

Just in case you weren't following what the other guys were saying ^

Answer 10

yup so
a=1
b=-1
c=20

Answer 11

c= -20*

Answer 12

I'm not sure if you'll need the fancy quadratic formula for this one :)
It looks like it will factor nicely!

Answer 13

yup but this is bit easy to explain lol

Answer 14

yes I understand changing it to 'u' but where do you go from there? I'm just unsure of how to simplify it?

Answer 15

@zepdrix @rvc

Answer 16

\[\large\rm u^2-1u-20=0\]Find factors of -20 which add to -1.

\(\rm -10\cdot2=-20\)
\(\rm -10+2\ne-1\)
Hmm these factors did not work out.

\(\rm -20\cdot1=-20\)
\(\rm -20+1\ne-1\)
Hmm these either...

Answer 17

itd -5 and 4

Answer 18

Try other factors of -20.

If you're really uncomfortable with factoring,
you can use the Quadratic Formula as Rvc suggested,\[\large\rm u=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]Where our quadratic is given by,\[\large\rm au^2+bu+c=0\]So we identify that:
\(\rm a=1\)
\(\rm b=-1\)
\(\rm c=-20\)

Answer 19

\[\large\rm u^2-1u-20=0\]Ok good. You can do it the shortcut way if you like,\[\large\rm (u-5)(u+4)=0\]Or you understand how we got to that point,
you can factor by grouping.
Rewrite -1u as the sum of -5u and 4u,\[\large\rm u^2-5u+4u-20=0\]Factor a u out of the first two terms,\[\large\rm u(u-5)+4u-20=0\]Factor a 4 out of the last two terms,\[\large\rm u(u-5)+4(u-5)=0\]Factor a (u-5) out of everything,\[\large\rm (u-5)(u+4)=0\]

Answer 20

From that point, apply your Zero-Factor Property,
setting each individual factor equal to zero,
(because a product can only be zero if one of the factors is zero),\[\large\rm u-5=0\qquad\qquad\qquad\qquad\qquad u+4=0\]And from this point,
you can undo your substitution,\[\large\rm e^x-5=0\qquad\qquad\qquad\qquad\qquad e^x+4=0\]and apply logs and all that fun stuff to solve for x.

Answer 21

ok so you would bring the -5 and 4 to the other side and then make it (ln)5 and (ln)-4 right?

Answer 22

Good good good.
Check to make sure that both of your solutions make sense.

Answer 23

\[\large\rm \ln(5)\approx 1.609\]\[\large\rm \ln(-4)=?\]

Answer 24

yeah ln(-4) us undefined so the answer is 1.609. thank you!

Answer 25

np