# logarithms help

• Apr 6th 2010, 06:02 PM
way123
logarithms help
hi, im powering through an advanced functions course right now by myself at home and could use some help.

1. Given that log 2 5 aproximately equals 2.3219
find an aproximation for log 2 20

2. 2log m + 3log m = 10 where there is no base on the logarithm.

• Apr 6th 2010, 06:14 PM
skeeter
Quote:

Originally Posted by way123
hi, im powering through an advanced functions course right now by myself at home and could use some help.

1. Given that log 2 5 aproximately equals 2.3219
find an aproximation for log 2 20

2. 2log m + 3log m = 10 where there is no base on the logarithm.

1. $\displaystyle \log_2(20) = \log_2(4 \cdot 5) = \log_2(4) + \log_2(5)$

2. no base is usually base 10 for beginners ...

$\displaystyle 2\log{m} + 3\log{m} = 5\log{m} = 10$

$\displaystyle \log{m} = 2$

change to an exponential equation to find $\displaystyle m$
• Apr 6th 2010, 06:21 PM
way123
wouldnt 5logm=10 be equivalent to log m^5=10
and in that case m is not equal to 2.
where am i going wrong?
• Apr 6th 2010, 06:26 PM
Stroodle
$\displaystyle log(x)$ usually means $\displaystyle log_{10}(x)$

and

$\displaystyle log_a(b)=c\Rightarrow a^c=b$
• Apr 6th 2010, 06:29 PM
skeeter
Quote:

Originally Posted by way123
wouldnt 5logm=10 be equivalent to log m^5=10
and in that case m is not equal to 2.
where am i going wrong?

m does not equal 2 ... $\displaystyle \log(m) = 2$

$\displaystyle m = 10^2$