1. ## 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.

2. 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 $\displaystyle m$

3. 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?

4. $\displaystyle log(x)$ usually means $\displaystyle log_{10}(x)$

and

$\displaystyle log_a(b)=c\Rightarrow a^c=b$

5. 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 ... $\displaystyle \log(m) = 2$

$\displaystyle m = 10^2$