Question

e^xe^(x+1)=1
Part 1: Combine the exponential expression to produce a single exponential expression.
Part 2: Convert the equation from part 1 to a logarithmic equation.
Part 3: Solve the equation.

Answer 1

So we have

\[\large e^xe^{x + 1} = 1\]

right?

Answer 2

Yea

Answer 3

We have exponents with the same base (e) so we just add the exponents

\[\large e^xe^{x + 1} \Rightarrow e^{x + x + 1} = e^{2x + 1}\]

make sense there?

Answer 4

2x+1=0,x=-1/2

Answer 5

So what we have is

\[\large e^{2x + 1} = 1\]

Answer 6

\[e ^{0}=1\]

Answer 7

Now for part 2, we want to change this to a logarithmic equation

Well our base being (e) we can take the natural log of both sides (ln)

\[\large \ln(e^{2x + 1}) = \ln(1)\]

Now remember \(\large \ln(e) = 1\) so we have

\[\large 2x + 1 = \ln(1)\]

Answer 8

Okay I understand that

Answer 9

Now we just solve

\[\large 2x = ln(1) - 1\]

We know ln(1) = 0 so

\[\large 2x = -1\]

And finally after dividing both sides by 2

\[\large x = -1/2\]

Answer 10

Okay great thanks so much for the help!

Answer 11

No problem!