1. ## Log simplification problem.

I am struggling to see the step below in a log problem.
I cannot see how you get to $\displaystyle \frac{3}{2}$

$\displaystyle \log_3 x=1-\frac{1}{2} \log_3 4$
$\displaystyle \log_3 x=1-\log_3 2$
$\displaystyle x = 3^{(1-\log_3 2)}$

$\displaystyle x = [3][3^{(-log_3 2)}]$

$\displaystyle x = \frac{3}{3^{(log_3 2)}}$

$\displaystyle x = \frac{3}{2}$

I don't understand how the last step is done?
How does that equal one and a half?

$\displaystyle \log_3 2 => 3^?=2$ confused.

2. Originally Posted by craigmain
I am struggling to see the step below in a log problem.
I cannot see how you get to $\displaystyle \frac{3}{2}$

$\displaystyle \log_3 x=1-\frac{1}{2} \log_3 4$
$\displaystyle \log_3 x=1-\log_3 2$
$\displaystyle x = 3^{(1-\log_3 2)}$

$\displaystyle x = [3][3^{(-log_3 2)}]$

$\displaystyle x = \frac{3}{3^{(log_3 2)}}$

$\displaystyle x = \frac{3}{2}$

I don't understand how the last step is done?
How does that equal one and a half?

$\displaystyle \log_3 2 => 3^?=2$ confused.
Conceptually, if $\displaystyle y=\log_3 2$, then $\displaystyle y$ is the exponent to which you need to raise $\displaystyle 3$ in order to get $\displaystyle 2$. So whatever $\displaystyle y$ is, you have $\displaystyle 3^y$, so by the definition of logs, you get out $\displaystyle 3^y=2$.

3. ## Just to clarify.

Are you saying:

we have:
$\displaystyle x=\frac{3}{3^{(\log_3 2)}}$

let $\displaystyle y=\log_3 2$ (1)

Therefore
$\displaystyle x=\frac{3}{3^y}$

but from (1)
$\displaystyle 3^y=2$

Therefore
$\displaystyle x = \frac{3}{2}$.

Sorry, but it seems like there's some circular logic here, so I want to be sure I am understanding you.

4. Right.