Solving Logarithm Problem: (-1/2)log2+(1/2)log1+(3/2)log4-(3/2)log4-(3/2)log3

[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 textbook. Could anyone please show me the intermediate steps between the two lines? Thank you.

 

[jedishrfu]

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)

 

[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.

 

[I like Serena]

Hi ZedCar!

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

 

[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]

 

[I like Serena]

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]

Good!

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

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?

 

[ZedCar]

Thanks very much I like Serena.

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

 

[ZedCar]

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

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

 

[I like Serena]

Cheers!

 

[I like Serena]

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

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.

 

[ZedCar]

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