Evaluating log

**BeSweeet** · September 28th 2009, 03:56 PM

Evaluate the function at the indicated value of $x$ .

I have to figure out what power times 8 equals 2. Anybody??

**VonNemo19** · September 28th 2009, 03:57 PM

Originally Posted by BeSweeet

Evaluate the function at the indicated value of $x$ .

I have to figure out what power times 8 equals 2. Anybody??

Do you know anything about fractional exponents? Because your answer is close.

Its REAL close.

**BeSweeet** · September 28th 2009, 04:01 PM

Originally Posted by VonNemo19

Do you know anything about fractional exponents? Because your answer is close.

Its REAL close.

Seriously? Because I don't know how to figure this one out :\.

**VonNemo19** · September 28th 2009, 06:02 PM

Originally Posted by BeSweeet

Seriously? Because I don't know how to figure this one out :\.

$x^{1/2}=\sqrt{x}$

$x^{1/3}=\sqrt[3]{x}$

$x^{1/4}=\sqrt[4]{x}$

$(27)^{1/3}=3$

Get the picture?

So.................. $(8)^{?}=2$ ?

**BeSweeet** · September 28th 2009, 07:09 PM

Not really

.

**mr fantastic** · September 28th 2009, 09:57 PM

Originally Posted by BeSweeet

Evaluate the function at the indicated value of $x$ .

I have to figure out what power times 8 equals 2. Anybody??

$\log_8 2 = x \Rightarrow 8^x = 2 \Rightarrow (2^3)^x = 2 \Rightarrow 2^{3x} = 2 \Rightarrow 2^{3x} = 2^1$ .... Solve for x.

**BeSweeet** · September 29th 2009, 07:45 AM

Ah, ok. We went through this in class and he got 3 :\.

log2^8
8=s^y
2^3=2^y
y=3

**mr fantastic** · September 29th 2009, 11:13 PM

Originally Posted by BeSweeet

Ah, ok. We went through this in class and he got 3 :\.

log2^8
8=s^y
2^3=2^y
y=3

That's not the question that was asked. Can you get the answer to the question that was asked by solving for x in my previous post?

Thread: Evaluating log