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

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

Mathematics
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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.

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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)
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?
e^x/e^2
Correct. So that means that our equation becomes?
2(e^x/e^2)-e^x=7
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)
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
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. \[ln[\frac{a}{b}] = (ln\ a) - (ln\ b)\]
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
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.. \[ln[e^{-2}] = -2\]
thanks for everything
Of course!
it said not real in calculator
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.
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

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