anonymous
  • anonymous
2(e^x-2)=(e^x)+7... Solve the equation
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Is that \[2e^{x-2} = e^x + 7\]?
anonymous
  • anonymous
ya
anonymous
  • anonymous
Ok, so first off, move all the \(e^x\) terms to one side of the equal sign.

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anonymous
  • anonymous
how do you do that
anonymous
  • anonymous
By adding or subtracting things from both sides.
anonymous
  • anonymous
For example, go ahead and subtract \(e^x\) from both sides and what do you get?
anonymous
  • anonymous
7=2(e^x-2)-(e^x)
anonymous
  • 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
  • anonymous
e^x/e^2
anonymous
  • anonymous
Correct. So that means that our equation becomes?
anonymous
  • anonymous
2(e^x/e^2)-e^x=7
anonymous
  • 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
  • anonymous
no sure how to right it now
anonymous
  • anonymous
u still there
anonymous
  • anonymous
\[\frac{2}{e^2}(e^x) - e^x = 7\] \[e^x(\frac{2}{e^2} - 1)= 7\]
anonymous
  • anonymous
Not sure why those fractions aren't showing up right..
anonymous
  • anonymous
But anyway, then you just divide by the second bit leaving only the \(e^x\) on the left side.
anonymous
  • anonymous
why don't we us ln
anonymous
  • anonymous
We will, next.
anonymous
  • anonymous
ok
anonymous
  • anonymous
You should have \[e^x = \frac{7}{(2e^{-2}) - 1}\] Then you take the ln of both sides.
anonymous
  • anonymous
so... x= ln7/(2ln -2) - ln1 not sure about denominator. Is that correct
anonymous
  • 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
  • anonymous
so... x=ln7-((2ln-2)-1) ?
anonymous
  • anonymous
i think i did something wrong bc u cnt have ln and then a negative. What woudit be then
anonymous
  • 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
  • anonymous
thanks for everything
anonymous
  • anonymous
Of course!
anonymous
  • anonymous
it said not real in calculator
anonymous
  • 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
  • anonymous
We must have missed something
anonymous
  • anonymous
Nope, that's the correct solution to this equation. It has no real soltions.
anonymous
  • anonymous
Solutions rather. Double check what I wrote originally.
anonymous
  • anonymous
what u originaly wrote is corect
anonymous
  • anonymous
must be no solutions thanks again

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