1. ## Easy Proof

Prove that $\displaystyle log_ax=\frac{lnx}{lna}$

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

2. Prove that $\displaystyle \log_a{x}=\frac{\ln x}{\ln a}$.

Let $\displaystyle \log_a{x} = y$.

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

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

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

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

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

Done!

01

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

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

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

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

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

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

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

Thus $\displaystyle \log_b{x} = \frac{\log{x}}{\log{b}}$.

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

Done!

01
Very Good! Unquestionably proven!