# Easy Proof

• Jun 19th 2009, 09:46 PM
VonNemo19
Easy Proof
Prove that $log_ax=\frac{lnx}{lna}$

The reason I'm asking this is because I was sleeping through logs. I'm now in calculus and in the proof of $\frac{d}{dx}(log_ax)=\frac{1}{(lna)x}$, they rewrite $log_ax$ as $\frac{lnx}{lna}$. I need to understand this so that I may sleep tonight. Thanx.
• Jun 19th 2009, 09:57 PM
yeongil
Prove that $\log_a{x}=\frac{\ln x}{\ln a}$.

Let $\log_a{x} = y$.

Write in exponential form:
$x = a^y$

Solve for y, but using natural logs. We do this by taking the natural log of both sides:
$\ln x = \ln (a^y)$

Use the property of logs to move the y over:
$\ln x = y \ln a$

Divide both sides by $\ln a$:
$\frac{\ln x}{\ln a} = y$

Substitute $\log_a{x}$ back in for y:
$\frac{\ln x}{\ln a} = \log_a{x}$

Done! (Wink)

01
• Jun 19th 2009, 10:00 PM
Prove It
Quote:

Originally Posted by VonNemo19
Prove that $log_ax=\frac{lnx}{lna}$

The reason I'm asking this is because I was sleeping through logs. I'm now in calculus and in the proof of $\frac{d}{dx}(log_ax)=\frac{1}{(lna)x}$, they rewrite $log_ax$ as $\frac{lnx}{lna}$. I need to understand this so that I may sleep tonight. Thanx.

If $x = b^a$ then $a = \log_b{x}$.

But we can evaluate $a$ in another way, using the formula

$\log{x^p} = p\log{x}$.

So $\log{x} = \log{b^a}$

$\log{x} = a\log{b}$

$a = \frac{\log{x}}{\log{b}}$.

Thus $\log_b{x} = \frac{\log{x}}{\log{b}}$.
• Jun 19th 2009, 10:05 PM
VonNemo19
Quote:

Originally Posted by yeongil
Substitute $\log_a{x}$ back in for y:
$\frac{\ln x}{\ln a} = \log_a{x}$

Done! (Wink)

01

Very Good! Unquestionably proven!