# Homework Help: Logarithm question

1. Jan 13, 2012

### ZedCar

(-1/2)log2 + (1/2)log1 + (3/2)log4 - (3/2)log4 - (3/2)log3

= (1/2)log[(4^3)/(2x3^3)]

The above is from a text book. Could anyone please show me the intermediate steps between the two lines? Thank you.

2. Jan 13, 2012

### Staff: Mentor

log A + log B = log AB

N log A = log (A^N)

similarly 2 log A + 0.5 log B = log (A^2) + log (B^0.5) = log (a^2 B^0.5)

3. Jan 13, 2012

### ZedCar

Thanks. I've been using the log laws, though I can't seem to get the same answer which is given in the book.

4. Jan 13, 2012

### I like Serena

Hi ZedCar!

Perhaps you can show us how you applied the log laws?

5. Jan 13, 2012

### ZedCar

Question is: (-1/2)log2 + (1/2)log1 + (3/2)log4 - (3/2)log4 - (3/2)log3

First I factored out 1/2

1/2 [-log2 + log1 + 3log4 - 3log4 - 3log3]

1/2 [-log(2x1) + 3log4 - 3log(4/3)]

1/2 [-log2 + 3log4 - 3log(4/3)]

1/2 [-log2 + 3log((4x3)/4))]

1/2 [3log((4x3)/4) - log2]

1/2 [3log((12x2)/4)]

1/2 [3log(24/4)]

1/2 [3log6]

6. Jan 13, 2012

### I like Serena

Good!

I'm afraid that this is not quite right.

You have not applied the priority rules for addition and subtraction properly.
They should be evaluated left-to-right, like this:

1/2 [((((-log2) + log1) + 3log4) - 3log4) - 3log3]

I've added parentheses to specify the order of evaluation.

In particular -log2 should be evaluated as -1 x log2 = log 2-1 = log(1/2).

Perhaps you can redo this step?

7. Jan 13, 2012

### ZedCar

Thanks very much I like Serena.

I've been able to figure it out now! Thank you.

8. Jan 13, 2012

### ZedCar

Is this the standard way of doing these? Start from the left and work right?

9. Jan 13, 2012

### I like Serena

Cheers!

10. Jan 13, 2012

### I like Serena

Yes.
If you have mixed additions and subtractions, you have to do them from left to right.

However, there is an alternative, which is like this:

1/2 [-log2 + log1 + 3log4 - 3log4 - 3log3]

1/2 [(-1 x log2) + log1 + (3 x log4) + (-3 x log4) + (-3 x log3)]

If you do it like this, you can add in an arbitrary order, but multiplication comes first.

11. Jan 13, 2012

### ZedCar

That's great. Thanks very much again for that!