# Logarithm Simplification Help. Any aid appreciated.

• Sep 19th 2010, 04:44 AM
BinaryBoy
Logarithm Simplification Help. Any aid appreciated.
Hi everyone,

I have been working with introductory logarithms recently, and I came across these two problems that I just could not solve. They are:

The questions are:

Simplify

Log3 (4 to the power of 2) + 3 log3 2

4 log3 2 – 2 log3 6 + 2

Any help simplifying these equations will be very much appreciated!

Thanks,

Nathaniel
• Sep 19th 2010, 05:42 AM
Prove It
$\log_3{(4^2)} + 3\log_3{2} = \log_3{(2^4)} + 3\log_3{2}$

$= 4\log_3{2} + 3\log_3{2}$

$= 7\log_3{2}$.
• Sep 19th 2010, 10:46 AM
Soroban
Hello, BinaryBoy!

Quote:

$\text{Simplify: }\;4\log_3(2) - 2\log_3(6) + 2$

$4\log_3(2) - 2\log_3(6) + 2 \;=\;\log_3\left(2^4\right) - \log_3\left(6^2\right) + 2$

. . . . . . . . . . . . . . . . . . $=\;\log_3(16)-\log_3(36) + 2$

. . . . . . . . . . . . . . . . . . $=\;\log_3\left(\frac{16}{36}\right) + 2$

. . . . . . . . . . . . . . . . . . $=\;\log_3\left(\frac{4}{9}\right) + 2$

. . . . . . . . . . . . . . . . . . $=\; \log_3(4) - \underbrace{\log_3(9)}_{\text{This is 2}} + 2$

. . . . . . . . . . . . . . . . . . $=\;\log_3(4) - 2 + 2$

. . . . . . . . . . . . . . . . . . $=\;\log_3(4)$

• Sep 19th 2010, 02:45 PM
BinaryBoy
Thank you so much Prove It, and Soroban! That was a huge help! :D

But, just to clear up one point...Soroban, could you please clarify why log3 (9) becomes 2? Is it because of the "logarithm of the base" rule?

Cheers,
• Sep 19th 2010, 04:03 PM
skeeter
Quote:

Originally Posted by BinaryBoy
... please clarify why log3 (9) becomes 2?

let $x = \log_3{9}$

then $3^x = 9$

$x = 2$

or ...

$\log_3{9} = \log_3(3^2) = 2\log_3{3} = 2 \cdot 1 = 2$
• Sep 20th 2010, 02:40 AM
BinaryBoy
Thanks for clarifying, skeeter. :)