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

- anonymous

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

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- anonymous

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

- anonymous

ya

- anonymous

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

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## More answers

- anonymous

how do you do that

- anonymous

By adding or subtracting things from both sides.

- anonymous

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

- anonymous

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

- anonymous

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?

- anonymous

e^x/e^2

- anonymous

Correct. So that means that our equation becomes?

- anonymous

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

- anonymous

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)

- anonymous

no sure how to right it now

- anonymous

u still there

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

why don't we us ln

- anonymous

We will, next.

- anonymous

ok

- anonymous

You should have
\[e^x = \frac{7}{(2e^{-2}) - 1}\]
Then you take the ln of both sides.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

thanks for everything

- anonymous

Of course!

- anonymous

it said not real in calculator

- anonymous

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.

- anonymous

We must have missed something

- anonymous

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

- anonymous

Solutions rather. Double check what I wrote originally.

- anonymous

what u originaly wrote is corect

- anonymous

must be no solutions thanks again

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