a basic logarithms question.

• Jun 15th 2007, 04:58 PM
beli3ve
a basic logarithms question.
$\displaystyle {25^{log_5 x}} = 5$

The small number right below the log, after it says it...

what is that and how would i find significance in that?

I have already converted the problem to look like the following:

$\displaystyle {log_{25}} 5 = log_5 x$.

I am looking to solve for X.
• Jun 15th 2007, 05:03 PM
Jhevon
Quote:

Originally Posted by beli3ve
$\displaystyle {25^{log_5 x}} = 5$

The small number right below the log, after it says it...

what is that and how would i find significance in that?

I have already converted the problem to look like the following:

$\displaystyle {log_{25}} 5 = log_5 x$.

I am looking to solve for X.

$\displaystyle 25^{ \log_5 x} = 5$

$\displaystyle \Rightarrow \left( 5^2 \right)^{ \log_5 x} = 5$

$\displaystyle \Rightarrow 5^{ 2 \log_5 x } = 5$

Now equate the powers, we get:

$\displaystyle 2 \log_5 x = 1$

can you take it from here?
• Jun 15th 2007, 05:07 PM
beli3ve
Quote:

Originally Posted by Jhevon

$\displaystyle 25^{ \log_5 x} = 5$

$\displaystyle \Rightarrow \left( 5^2 \right)^{ \log_5 x} = 5$

$\displaystyle \Rightarrow 5^{ 2 \log_5 x } = 5$

Now equate the powers, we get:

$\displaystyle 2 \log_5 x = 1$

can you take it from here?

i am unsure.
from here, i would divide the two to the other side, making it.

$\displaystyle log_5 x = 1/2$.

that is an option on my paper, but i am still confused to the significance of what the 5 is?
actually, as I was just sitting here thinking, is that the number it would represent that was sharing the powers?
• Jun 15th 2007, 05:10 PM
Jhevon
Quote:

Originally Posted by beli3ve
i am unsure.
from here, i would divide the two to the other side, making it.

$\displaystyle log_5 x = 1/2$.

that is an option on my paper, but i am still confused to the significance of what the 5 is?
actually, as I was just sitting here thinking, is that the number it would represent that was sharing the powers?

again, remember the rule i told you about.

If $\displaystyle \log_a b = c$, then $\displaystyle a^c = b$

So we have $\displaystyle 2 \log_5 x = 1$ .....you divided both sides by 2, which is fine, but is not what i would have done

$\displaystyle \Rightarrow \log_5 x = \frac {1}{2}$

$\displaystyle \Rightarrow 5^{ \frac {1}{2}} = x$ .....i applied the law above

$\displaystyle \Rightarrow x = \sqrt {5}$
• Jun 15th 2007, 05:16 PM
beli3ve
Quote:

Originally Posted by Jhevon
again, remember the rule i told you about.

If $\displaystyle \log_a b = c$, then $\displaystyle a^c = b$

So we have $\displaystyle 2 \log_5 x = 1$ .....you divided both sides by 2, which is fine, but is not what i would have done

$\displaystyle \Rightarrow \log_5 x = \frac {1}{2}$

$\displaystyle \Rightarrow 5^{ \frac {1}{2}} = x$ .....i applied the law above

$\displaystyle \Rightarrow x = \sqrt {5}$

the

$\displaystyle x = \sqrt {5}$

makes little sense to me because

$\displaystyle x = 5^{1/2}$ & $\displaystyle x = \sqrt {5}$

are not the same thing to me.
to me, it could only the square root if

$\displaystyle x = 5^{2}$
• Jun 15th 2007, 05:26 PM
Jhevon
Quote:

Originally Posted by beli3ve
the

$\displaystyle x = \sqrt {5}$

makes little sense to me because

$\displaystyle x = 5^{1/2}$ & $\displaystyle x = \sqrt {5}$

are not the same thing to me.
to me, it could only the square root if

$\displaystyle x = 5^{2}$

ummm, $\displaystyle 5^{ \frac {1}{2}}$ is $\displaystyle \sqrt {5}$

see http://www.mathhelpforum.com/math-he...tial-form.html

and look at your statement again. you said x = 5^2 means x is the squareroot of 5. that makes no sense. if x = 5^2, it means x is the square of 5, since if we square 5, we get 5^2 which is equal to x
• Jun 15th 2007, 05:47 PM
beli3ve
Quote:

Originally Posted by Jhevon
ummm, $\displaystyle 5^{ \frac {1}{2}}$ is $\displaystyle \sqrt {5}$

see http://www.mathhelpforum.com/math-he...tial-form.html

and look at your statement again. you said x = 5^2 means x is the squareroot of 5. that makes no sense. if x = 5^2, it means x is the square of 5, since if we square 5, we get 5^2 which is equal to x

that makes a lot more sense to me now.
i do see my errors.
again, thank you.
• Jun 15th 2007, 05:53 PM
Jhevon
Quote:

Originally Posted by beli3ve
that makes a lot more sense to me now.
i do see my errors.
again, thank you.

ok. keep practicing