Question

Solve for x:
2e^(x-2) = e^x + 7

Answer 1

Note that
\[\large\begin{aligned} 2e^{x-2}=e^x+7 &\implies 2e^{-2}e^{x} = e^x+7\\ &\implies2e^{-2}e^x-e^x = 7\\ &\implies e^x(2e^{-2}-1)=7\\ &\implies e^x=\frac{7}{2e^{-2}-1} \end{aligned}\]
So, what is \(x\)? :-)

Answer 2

how did you get from 2e^(x-2) to 2e^-2 e^x?

Answer 3

i got math error

Answer 4

I used the property of exponents \(\large a^{b+c} = a^ba^c\), Thus, \(\large e^{x-2} = e^{x+(-2)} = e^xe^{-2}\).

Does this clarify things? :-)

Answer 5

right!!! :D didnt think of that

Answer 6

@ChristopherToni

Answer 7

That is correct. Because of this, you should see that there is no solution.

In particular, the reason why there isn't a solution is due to the fact that \(\large 2e^{-2}-1<0\) and hence when you take natural logs of both sides of \(\large e^x = \dfrac{7}{2e^{-2}-1}\), you get no solution (because \(\large \ln x\) is only defined for \(\large x>0\)).

Does this clarify things? :-)

Answer 8

i think it does :D thanks. by the way, I like how you always put "does this clarify things? :-)" at the end.

Answer 9

I try to be courteous whenever possible. XD