# Thread: Can someone help me solve this logerithm

1. ## Can someone help me solve this logerithm

Solve the equation

(e^4) * e^x2=e^12

I come up with x^3=12 ,then x^3-12=0.

I know I'm wrong, but I can't see a way to simplify x^3

Thanks

Jason

2. Originally Posted by Darkhrse99
Solve the equation

(e^4) * e^x2=e^12

I come up with x^3=12 ,then x^3-12=0.

I know I'm wrong, but I can't see a way to simplify x^3

Thanks

Jason
?? did you type the question correctly? how do you end up with x^3 anywhere?

is it supposed to be $\displaystyle e^4 e^{x^2} = e^{12}$?

3. Originally Posted by Jhevon
?? did you type the question correctly? how do you end up with x^3 anywhere?

is it supposed to be $\displaystyle e^4 e^{x^2} = e^{12}$?
I left out the power of X after (e^4) . I came up with x^3 when i combined x^2 and x. The last part is right. but it's (e^4)^x. It's e the the 4th power times the power of x.

4. Originally Posted by Darkhrse99
Yes that is the equation. I came up with x^3 when i combined x^2 and x.
that's what i'm saying. what x? there is no x here. the only x-term is the x^2

we have: $\displaystyle e^4e^{x^2} = e^{x^2 + 4} = e^{12}$

now equate the powers and continue

5. Originally Posted by Jhevon
that's what i'm saying. what x? there is no x here. the only x-term is the x^2

we have: $\displaystyle e^4e^{x^2} = e^{x^2 + 4} = e^{12}$

now equate the powers and continue
It's (e^4)^x * e^x^2=e^12

6. Originally Posted by Darkhrse99
It's (e^4)^x * e^x^2=e^12
ok, one more time for clarity's sake. it is $\displaystyle \left( e^4 \right)^x e^{x^2} = e^{12}$?

read carefully this time and make sure this is the problem

7. Originally Posted by Jhevon
ok, one more time for clarity's sake. it is $\displaystyle \left( e^4 \right)^x e^{x^2} = e^{12}$?

read carefully this time and make sure this is the problem
Yes that is the correct problem. Sorry about the confusion. I was editing while you were posting.

8. Originally Posted by Darkhrse99
Yes that is the correct problem. Sorry about the confusion. I was editing while you were posting.
okay, so you should know two things here, x^3 will still not come into play:

$\displaystyle \left( x^a \right)^b = x^{ab}$ ....................(1)

$\displaystyle x^a \cdot x^b = x^{a + b}$ ................(2)

now on to the problem

$\displaystyle \left( e^4 \right)^x e^{x^2} = e^{12}$

$\displaystyle \Rightarrow e^{4x}e^{x^2} = e^{12}$ ................by (1)

$\displaystyle \Rightarrow e^{x^2 + 4x} = e^{12}$ ................by (2)

equating the powers we get

$\displaystyle x^2 + 4x = 12$

now conitnue

9. I get

x^2+4x=12
x^2+4x-12=0
(x-2) (x+6)=0
x=2 x=-6 Correct?

10. Originally Posted by Darkhrse99
I get

x^2+4x=12
x^2+4x-12=0
(x+4) (x-3)=0
x=-4 x=3 Correct?
no

expand (x + 4)(x - 3), what do you get?

i guarantee you do not get x^2 + 4x - 12

11. x=-6 x=2 My hands aren't following my brain tonight.uugghh

12. Originally Posted by Darkhrse99
x=-6 x=2 My hands aren't following my brain tonight.uugghh
yes, those are the solutions to the quadratic. make sure to check both solutions in the original equation to make sure they work

13. Ok thanks for the help and sorry for the confusion. What program are you guys workin with to write out problems.?

14. Originally Posted by Darkhrse99
Ok thanks for the help and sorry for the confusion.
you're welcome

and that's ok. good luck with your class