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

ya

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

how do you do that

By adding or subtracting things from both sides.

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

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

e^x/e^2

Correct. So that means that our equation becomes?

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

no sure how to right it now

u still there

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

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

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

why don't we us ln

We will, next.

ok

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

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

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

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

thanks for everything

Of course!

it said not real in calculator

We must have missed something

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

Solutions rather. Double check what I wrote originally.

what u originaly wrote is corect

must be no solutions thanks again