Can someone help me solve this logerithm

**Darkhrse99** · November 25th 2007, 07:21 PM

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

**Jhevon** · November 25th 2007, 07:23 PM

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 $e^4 e^{x^2} = e^{12}$ ?

**Darkhrse99** · November 25th 2007, 07:28 PM

Originally Posted by Jhevon

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

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

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.

**Jhevon** · November 25th 2007, 07:30 PM

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: $e^4e^{x^2} = e^{x^2 + 4} = e^{12}$

now equate the powers and continue

**Darkhrse99** · November 25th 2007, 07:38 PM

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: $e^4e^{x^2} = e^{x^2 + 4} = e^{12}$

now equate the powers and continue

It's (e^4)^x * e^x^2=e^12

**Jhevon** · November 25th 2007, 07:40 PM

Originally Posted by Darkhrse99

It's (e^4)^x * e^x^2=e^12

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

read carefully this time and make sure this is the problem

**Darkhrse99** · November 25th 2007, 07:42 PM

Originally Posted by Jhevon

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

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.

**Jhevon** · November 25th 2007, 07:47 PM

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:

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

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

now on to the problem

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

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

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

equating the powers we get

$x^2 + 4x = 12$

now conitnue

**Darkhrse99** · November 25th 2007, 07:54 PM

I get

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

**Jhevon** · November 25th 2007, 07:56 PM

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

**Darkhrse99** · November 25th 2007, 07:58 PM

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

**Jhevon** · November 25th 2007, 08:04 PM

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

**Darkhrse99** · November 25th 2007, 08:06 PM

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

**Jhevon** · November 25th 2007, 08:07 PM

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

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