# logarithm problem

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• May 8th 2009, 10:06 PM
meiyukichan
logarithm problem
Show that for any positive numbers $\displaystyle x$ and $\displaystyle a$ (with $\displaystyle x$ not equal to $\displaystyle 1$) $\displaystyle \log_{x^2} a = \frac{1}{2} \log_{x} a$.

Also, generalize the assertion in the problem above to powers other than $\displaystyle x^2$.

Thank you (: (Bow)
• May 8th 2009, 10:08 PM
Rapha
Hi.

Quote:

Originally Posted by meiyukichan
Show that for any positive numbers $\displaystyle x$ and $\displaystyle a$ (with $\displaystyle x$ not equal to $\displaystyle 1$) $\displaystyle \log_{x^2} a = \frac{1}{2} \log_{x} a$.

Also, generalize the assertion in the problem above to powers other than $\displaystyle x^2$.

Do you know this rule: $\displaystyle log_(y)x = \frac{ln(x)}{ln(y)}$ (rule1)?

In this case $\displaystyle \log_{x^2} a = \frac{ln(a)}{ln(x^2)} = \frac{1}{2}\frac{ln(a)}{ln(x)}$

because (rule2) $\displaystyle log(x^n) = n log(x)$

Using (rule1)
$\displaystyle 1/2 * log_x(a)$

Do you know these rules or do you have to proove them?
• May 9th 2009, 12:52 AM
meiyukichan
Quote:

Originally Posted by Rapha
Hi.

Do you know this rule: $\displaystyle log_(y)x = \frac{ln(x)}{ln(y)}$ (rule1)?

In this case $\displaystyle \log_{x^2} a = \frac{ln(a)}{ln(x^2)} = \frac{1}{2}\frac{ln(a)}{ln(x)}$

because (rule2) $\displaystyle log(x^n) = n log(x)$

Using (rule1)
$\displaystyle 1/2 * log_x(a)$

Do you know these rules or do you have to proove them?

Now that you mentioned it, I recognize these rules and it helps with the first problem. I understand it now.
Thank you! (:
• May 9th 2009, 03:51 AM
HallsofIvy
The rule follows from:

If $\displaystyle \log_{x^2} a= y$, then $\displaystyle a= (x^2)^y= x^{2y}$ so $\displaystyle \log_x a= 2y$ and $\displaystyle y= \log_{x^2} a= \frac{1}{2}\log_x a$