Question

Solve for x : e^x e^(x+1) = 1

Answer 1

i think its xe^2x+1

Answer 2

but i could be wrong.

Answer 3

But how would I solve for x? I don't understand the solving for x when there are more than one x in the equation ha

Answer 4

@aylorterb You should use the fact that anything to the zero power is just 1, so e^0=1. That will let you solve maybe?

Answer 5

@Kainui how did you get e^0? which e would that be?

Answer 6

Ahhh, I'm saying you can turn your equation from: \[\LARGE e^x e^{(x+1)} = 1\] into \[\LARGE e^x e^{(x+1)} =e^0\] since \[\LARGE e^0=1\]

Sorry for the confusion.

Answer 7

Okay! No need to be sorry! Thank you!

Answer 8

yes just mentioning tha e^0 comes from:

a=e^(ln a) is the rule, and you know that ln(1)=0
So, 1=e^(ln 1) = e^0

then equate the exponents, after the second line that Kainui wrote.

Answer 9

So.. ln (e^x) ln (e^(x+1)) = ln 1
x (x+1) = ln 1?

Answer 10

Not quite, but you're on the right path. The rule for a product is \[\LARGE \ln(ab)=\ln(a)+\ln(b)\] So when you separate the two e^x and e^(x+1) terms you should add them, not multiply.

Answer 11

okay so it would be x+ (x+1) = ln 1 ?
And then I could subtract 1 from both sides, and then divide by 2?

Answer 12

YES!
basically: \(\large\color{slate}{ e^xe^{x+1}=1 }\)

\(\large\color{slate}{ e^{x+x+1}=1 }\)

\(\large\color{slate}{ e^{x+x+1}=e^{\ln(1)} }\)

\(\large\color{slate}{ e^{2x+1}=e^{0} }\)

\(\large\color{slate}{ 2x+1=0 }\)

and then solve for \(\large\color{slate}{ x }\)...

Answer 13

Thank you guys so much!

Answer 14

yes, just a very good things to know: \(\Large\color{teal}{ a=e^{\ln(a)} }\) \(\LARGE\color{slate}{ ... }\) yw!