# Law of logarithm inquiry

• Jan 20th 2011, 05:30 AM
danzig
Law of logarithm inquiry
I'm curious as to how the answers for these came up.

For problem 1, given that log2 7 is approximately equal to 2.8074, find an approximation for each logarithm. Solve log2 3 square root of 7.

For problem 2, given that log2 5 is approximately equal to 2.3219, find an approximation for each logartihm.

http://img810.imageshack.us/img810/8651/logw.th.png

If anyone could shed some light on these problems, I dont understand how in problem 1 in the solution is not the sum of two logarithms like in problem 2, and how in problem 1 uses multiplication and problem 2 uses addition. I just need a bit of guidance, i'm a highschool student and I would like to understand this for upcoming exams.
• Jan 20th 2011, 05:34 AM
Prove It
For the first, $\displaystyle \log_2{\sqrt[3]{7}} = \log_2{(7^{\frac{1}{3}})} = \frac{1}{3}\log_2{7}$...

For the second, $\displaystyle \log_2{20} = \log_2{(4\cdot 5)} = \log_2{4} + \log_2{5} = \log_2{2^2} + \log_2{5}$...
• Jan 20th 2011, 05:46 AM
danzig
Thanks for the quick reply, Prove It

I just dont understand why it says "Given that log2 7 is approximately equal to 2.8074, find an approximation for each logarithm. Solve log2 3 square root of 7" , if the solution does not include the sum of two logarithms?
• Jan 20th 2011, 05:50 AM
Prove It
It's counting all logarithms from both questions...
• Jan 21st 2011, 06:20 AM
HallsofIvy
Quote:

Originally Posted by danzig
Thanks for the quick reply, Prove It

I just dont understand why it says "Given that log2 7 is approximately equal to 2.8074, find an approximation for each logarithm. Solve log2 3 square root of 7" , if the solution does not include the sum of two logarithms?

I don't understand why you think it should include the sum of two logarithms. Not every problem in the world involves adding!

Whoever gave you these problems expects that you already know the "laws of logarithms" that you mention in the title but not in your post:

$log(ab)= log(a)+ log(b)$
$log(a/b)= log(a)- log(b)$
$log(a^b)= b log(a)$.
$log(\sqrt[b]{a})= log(a^{1/b})= \frac{log(a)}{b}$