Expert: Paul Klarreich - 2/27/2007

Question
2/3log27+2log2-2log3;

4in2+1/2in4-in8;

2log1/3-3log1/2

Answer
Questioner:   pearl
Category:  Advanced Math
 
Subject:  logarithms
Question:  2/3log27+2log2-2log3;

4in2+1/2in4-in8;

2log1/3-3log1/2
..................................
Hi, Pearl,

PLEASE, when you send a problem, include instructions.  When you write:

4in2+1/2in4-in8

what are you supposed to do with it?  Every textbook (I assume that's where you got these) will have problem sets that say:

In 7-16, do such and such:

7.....
8.....
...
16.....

The 'such and such' is part of the problem, and if you leave it out, I have to guess, and if I guess wrong, that doesn't do you any good.

I'm guessing here that the instructions say:

USE BASIC LOGARITHM PRINCIPLES TO SIMPLIFY AND WRITE AS A SINGLE LOGARITHM OF A NUMBER:

(2/3)log 27 + 2 log 2 - 2 log 3

(2/3) log 3^3 + 2 log 2 - 2 log 3

(2/3)(3)log 3 + 2 log 2 - 2 log 3

2 log 3 + 2 log 2 - 2 log 3

2 log 2 = log 4
......................

4in2+1/2in4-in8;

I assume that's  LN, not IN.

4 ln 2 + 1/2 ln 4 - ln 8

4 ln 2 + 1/2 ln 2^2 - ln 2^3

4 ln 2 + ln 2 - 3 ln 2

2 ln 2 = ln 4
........................

2 log(1/3) - 3 log(1/2)
 
- 2 log 3 + 3 log 2

3 log 2 - 2 log 3
log 2^3 - log(3^2)

log 8 - log 9

log (8/9)