# Exponent rules with logarithms

• Mar 4th 2008, 08:52 PM
Exponent rules with logarithms
Hello,

I'm having trouble remembering a basic rule.

The question is:
(e^x)^2 = 3

My next step is:

ln(e^x)^2 = ln(3)

To my understanding, the ln and e knock each other out bringing down the x and 2. My question is, do I add them to make:

x+2 = ln(3) ? (Thinking) I'm not sure about the left side of that equation.

Thanks for any help!
• Mar 4th 2008, 09:13 PM
Here's what I got...so unsure about it...

2x = ln(3)
x=1/2 ln (3)
x= 0.549
• Mar 4th 2008, 10:53 PM
earboth
Quote:

Here's what I got...so unsure about it...

2x = ln(3)
x=1/2 ln (3)
x= 0.549

Your calculations are OK. I personally prefer:

$x=\frac12 \cdot \ln(3) = \ln(\sqrt{3})$
• Mar 4th 2008, 10:56 PM
earboth
Quote:

Hello,

I'm having trouble remembering a basic rule.

The question is:
(e^x)^2 = 3

My next step is:

ln((e^x)^2) = ln(3)

To my understanding, the ln and e knock each other out bringing down the x and 2. My question is, do I add them? No

$(e^x)^2 = 3~\iff~ e^{2x} = 3~\implies~2x = \ln(3)$
• Mar 5th 2008, 05:44 AM
Soroban

Quote:

Solve for $x\!:\;\;(e^x)^2 \:= \:3$
Take the square root of both sides: . $e^x \:=\:\sqrt{3}$

Take logs: . $\ln(e^x) \:=\:\ln(\sqrt{3}) \quad\Rightarrow\quad x\cdot\ln(e) \:=\:\ln(\sqrt{3})$

Since $\ln(e) = 1$, we have: . $x \:=\:\ln(\sqrt{3})$

• Mar 5th 2008, 06:09 AM
colby2152
Quote:

Hello,

I'm having trouble remembering a basic rule.

The question is:
(e^x)^2 = 3

My next step is:

ln(e^x)^2 = ln(3)

To my understanding, the ln and e knock each other out bringing down the x and 2. My question is, do I add them to make:

x+2 = ln(3) ? (Thinking) I'm not sure about the left side of that equation.

Thanks for any help!

You are fine until this...

$ln(e^x)^2 = ln(3)$

Rules of logs say that the power ("2" in this case) gets brought down as a coefficient...

$2ln(e^x) = ln(3)$

Now, cancel out the natural log with base e

$2x = ln(3)$

Solve for x...

$x = \frac{ln(3)}{2} \Rightarrow ln(3^{\frac{1}{2}}) = ln(\sqrt{3})$
• Mar 5th 2008, 08:18 AM