I think this one is a bit different, so help solving for x? ._.

(Posting the equation in comments)

Question

I think this one is a bit different, so help solving for x? ._.

(Posting the equation in comments)

anonymous · Answer

$$e ^{x}e ^{(x+1)}=1$$

anonymous · Answer

I know e goes with natural logarithms... need help isolating x though, since it's an exponent in this case...

anonymous · Answer

try multiplying both sides by the ln function

anonymous · Answer

well, ln1 is 0, and lne would cancel the e out.... I think, right?

anonymous · Answer

yes thats right

anonymous · Answer

So... x(x+1) = 0?

anonymous · Answer

you cant take the ln of zero so after you take the first ln to get rid of 1 e, you cant go any further so x would approach infinity. thats what i think but i may be wrong. do you know the answer?

ranga · Answer

$$\Large a^m * a^n = a^{m+n}$$

anonymous · Answer

Realize that $$e^x*e^(x+1) =  e^(2x+1)$$

anonymous · Answer

So your equation is simplified to e^(2x+1) = 1

anonymous · Answer

Ohh okay.. so if I found the natural logarithms of both sides it would leave me with 2x + 1 = 0?

anonymous · Answer

bingo

ranga · Answer

yes.

anonymous · Answer

So ultimately x  = -1/2

ranga · Answer

yes.

isaiah.feynman · Answer

Oh no! I missed a nice exponential equation to solve!

ranga · Answer

you can put it back in the original equation and verify it fpr yourself.

anonymous · Answer

Yay! And awh Isaiah I'm sure I'll have more to come eventually xD I'll tag you in them if I have anymore questions.