1. ## Log simplification problem.

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

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

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

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

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

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

$\log_3 2 => 3^?=2$ 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 $\frac{3}{2}$

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

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

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

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

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

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

3. ## Just to clarify.

Are you saying:

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

let $y=\log_3 2$ (1)

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

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

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

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

4. Right.