# Logarithm help!

• May 4th 2008, 07:59 PM
Croyt
Logarithm help!
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

• May 4th 2008, 08:43 PM
Reckoner
Hello Croyt!

Quote:

Originally Posted by Croyt
1. $\displaystyle 3\log_2(x) + \frac12\log_2(x) - 3\log_2(x) - 5$

Quote:

Originally Posted by Croyt
2. If $\displaystyle \log_b(3) = 0.56$ and $\displaystyle \log_b(2) = 0.36$ evaluate $\displaystyle \log_b\sqrt{\frac23}$

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

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

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

Quote:

Originally Posted by Croyt
3. Solve the equation for $\displaystyle x$. Give an exact answer and a four place decimal approximation

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

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

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

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

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

Quote:

Originally Posted by Croyt
4. Solve

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

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