Logarithm help!

**Croyt** · May 4th 2008, 07:59 PM

Hey guys, I appreciate the help on these logs and I ran into another snag

I tried to use logarit for these but I keep getting the wrong answer

1. 3log(base 2) x + 1/2 log(base 2) x - 3log (base 2) x - 5

2. If log (base b) 3 = 0.56 and log (base b) 2 = 0.36 evaluate...

log (base b) square root of 2/3

3. Solve the equation for x. Give an exact answer and a four place decimal approximation

log (2x+1) = -0.9

4. Solve

log (base 2) x + log (base 2) (3x-2) = 3

PLEASE PLEASE PLEASE HELP!!! Answers are greatly appreciated!

**Reckoner** · May 4th 2008, 08:43 PM

Hello Croyt!

Originally Posted by Croyt

1. $3\log_2(x) + \frac12\log_2(x) - 3\log_2(x) - 5$

Originally Posted by Croyt

2. If $\log_b(3) = 0.56$ and $\log_b(2) = 0.36$ evaluate $\log_b\sqrt{\frac23}$

Here are some basic rules that will help you:

1. $a\log x=\log x^a$

2. $\log xy=\log x + \log y$

3. $\log\frac xy=\log x - \log y$

Originally Posted by Croyt

3. Solve the equation for $x$ . Give an exact answer and a four place decimal approximation

$\log(2x+1) = -0.9$

Are you dealing with base $\text e$ , or base 10? In either case, the inverse operation of the logarithm is exponentiation. So, for example,

$\ln(2x+1)=-0.9$

$\Rightarrow \text e^{\ln(2x+1)}=\text e^{-0.9}$

$\Rightarrow 2x+1=\text e^{-0.9}$

Originally Posted by Croyt

4. Solve

$\log_2(x) + \log_2(3x-2) = 3$

Use the properties I listed above to simplify, and then solve it like number 3.

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