1. Simplifying and/or Computing problems

Hi I need help simplifing these problems.

1. 2ln $sqrt(e)$
= $sqrt(3)^2$
= 3

2 $^Logx$ = 2^log4 x
= 4 $^3x$

3. tan $^-1$(1/ $sqrt(3)$
= pi/3

are these right?

2. Originally Posted by killasnake
1. 2ln $sqrt(e)$
= $sqrt(3)^2$
= 3
Where did the 3 come from?

$2~ln(\sqrt{e}) = 2~ln \left ( e^{1/2} \right ) = 2~ \frac{1}{2} \cdot ln(e) = 1 \cdot 1 = 1$

-Dan

3. Originally Posted by killasnake
2 $^Logx$ = 2^log4 x
= 4 $^3x$
I presume "Log" is $log_{10}$?

Use the change of base formula:
$log_{10}(x) = \frac{log_2(x)}{log_2(10)}$

So
$2^{log_{10}(x)} = 2^{\frac{log_2(x)}{log_2(10)}}$

$= \left ( 2^{log_2(x)} \right )^{1/
log_2(10)}$

$= x^{1/log_2(10)} \approx x^{0.30103}$

-Dan

4. Originally Posted by killasnake
3. tan $^-1$(1/ $sqrt(3)$
= pi/3
Close.
$tan^{-1} \left ( \frac{1}{\sqrt{3}} \right ) = \frac{\pi}{6}$

-Dan

5. Originally Posted by topsquark
Where did the 3 come from?

$2~ln(\sqrt{e}) = 2~ln \left ( e^{1/2} \right ) = 2~ \frac{1}{2} \cdot ln(e) = 1 \cdot 1 = 1$

-Dan
oh crap i just realised my problem... the 3 was suppose to be an e