# Thread: a basic logarithms question.

1. ## a basic logarithms question.

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

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:

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

I am looking to solve for X.

2. Originally Posted by beli3ve
${25^{log_5 x}} = 5$

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:

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

I am looking to solve for X.

$25^{ \log_5 x} = 5$

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

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

Now equate the powers, we get:

$2 \log_5 x = 1$

can you take it from here?

3. Originally Posted by Jhevon

$25^{ \log_5 x} = 5$

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

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

Now equate the powers, we get:

$2 \log_5 x = 1$

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

$log_5 x = 1/2$.

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?

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

$log_5 x = 1/2$.

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

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

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

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

$\Rightarrow x = \sqrt {5}$

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

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

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

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

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

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

$x = \sqrt {5}$

makes little sense to me because

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

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

$x = 5^{2}$

6. Originally Posted by beli3ve
the

$x = \sqrt {5}$

makes little sense to me because

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

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

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

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

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

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.

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