Question

can someone help me solve for x
e^x e^(x+1)=1

Answer 1

Is this all of your question?

Answer 2

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

Answer 3

yes

Answer 4

What do you think it is?

Answer 5

Recall your exponent rule:\[\large\rm a^b\cdot a^c\quad=a^{b+c}\]

Answer 6

Do you see how that will help us on the left side of the equation?

Answer 7

yes

Answer 8

i think it might be 0

Answer 9

For x? Hmm no.

Answer 10

Applying the rule to the left side gives us:\[\large\rm e^{x+(x+1)}=1\]\[\large\rm e^{2x+1}=1\]Then apply natural log to each side.
Or rewrite 1 as e^0.

Whichever method you prefer

Answer 11

\[\large\rm e^{2x+1}=e^0\]

Answer 12

would we cancel out e

Answer 13

Bases are equivalent, both e,
so the exponents must be equivalent as well, ya? :)
\[\large\rm 2x+1=0\]

Sure, you can think of it as cancelling out, ya

Answer 14

would we get -0.5

Answer 15

Yes :)

Answer 16

yay! so the answer is x=-0.5

Answer 17

yay good job

Answer 18

Thank you so much!