Question

2(e^x-2)=(e^x)+7... Solve the equation

Answer 1

Is that \[2e^{x-2} = e^x + 7\]?

Answer 2

ya

Answer 3

Ok, so first off, move all the \(e^x\) terms to one side of the equal sign.

Answer 4

how do you do that

Answer 5

By adding or subtracting things from both sides.

Answer 6

For example, go ahead and subtract \(e^x\) from both sides and what do you get?

Answer 7

7=2(e^x-2)-(e^x)

Answer 8

Good. Now lets rewrite that \(e^{x-2}\) as the product of two powers of e.
Recall that \[a^b * a^c = a^{b+c}\]

So what can we rewrite \[e^{x-2}\] as?

Answer 9

e^x/e^2

Answer 10

Correct. So that means that our equation becomes?

Answer 11

2(e^x/e^2)-e^x=7

Answer 12

Right. Now since both terms on the left side have an \(e^x\) factor, we can factor it out in front of the sum.

a*b + a*c = a(b+c)

Answer 13

no sure how to right it now

Answer 14

u still there

Answer 15

\[\frac{2}{e^2}(e^x) - e^x = 7\]
\[e^x(\frac{2}{e^2} - 1)= 7\]

Answer 16

Not sure why those fractions aren't showing up right..

Answer 17

But anyway, then you just divide by the second bit leaving only the \(e^x\) on the left side.

Answer 18

why don't we us ln

Answer 19

We will, next.

Answer 20

ok

Answer 21

You should have

\[e^x = \frac{7}{(2e^{-2}) - 1}\]

Then you take the ln of both sides.

Answer 22

so... x= ln7/(2ln -2) - ln1
not sure about denominator. Is that correct

Answer 23

No, you have to take the ln of the whole thing, not the ln of the top divided by the ln of the bottom. If you want to break it up further you can use the property of log of a quotient is subtraction of the log.

\[ln[\frac{a}{b}] = (ln\ a) - (ln\ b)\]

Answer 24

so... x=ln7-((2ln-2)-1) ?

Answer 25

i think i did something wrong bc u cnt have ln and then a negative. What woudit be then

Answer 26

It should be:
\[x = (ln\ 7) - (ln [2e^{-2} - 1])\]
It cannot be simplified further because we can't do anything with the log of a sum. But you can plug it into your calculator if you like.

I think you may have tried to do \((ln\ e^{-2})\) which isn't legal in this case, but is legal in other situations. It's not equal to \(ln -2\) though..

\[ln[e^{-2}] = -2\]

Answer 27

thanks for everything

Answer 28

Of course!

Answer 29

it said not real in calculator

Answer 30

Hrmph.. That's true actually. \(2e^{-2}\) is less than 1, so when you subtract 1 from it you'll have a negative number and you cannot take the ln of a negative.

Answer 31

We must have missed something

Answer 32

Nope, that's the correct solution to this equation. It has no real soltions.

Answer 33

Solutions rather. Double check what I wrote originally.

Answer 34

what u originaly wrote is corect

Answer 35

must be no solutions thanks again