[SPLIT] how does log_b (3b^2)/(125) equal this...?

[Lime]

Here's another juicy logarithm problem to throw your way.

log_b (3b^2)/(125) = log_b (3b^2) - log_b (125)

I get this, based on the #1 Property of Logarithms (according to my textbook)

= log_b (3) + log_b (b^2) - log_b (5^3) ... :?:

Can someone explain how you get all three of those "things"?

 

[stapel]

If they gave you the "turn division inside the log into subtraction outside the log" rule, then they should have given you the corresponding multiplication-to-addition rule. Just apply that.

Eliz.

P.S. What is 5³?

 

[Lime]

If they gave you the "turn division inside the log into subtraction outside the log" rule, then they should have given you the corresponding multiplication-to-addition rule. Just apply that.

Eliz.

P.S. What is 5³?

Can't see it anywhere. Can you explain it?

And yeah, it's 125. I just didn't want to assume anything.

 

[stapel]

If they gave you the "turn division inside the log into subtraction outside the log" rule, then they should have given you the corresponding multiplication-to-addition rule. Just apply that.

Can't see it anywhere. Can you explain it?

It's very odd that they would have given you the one rule, but not the two others....

For a complete explanation with worked examples, try this lesson on the log rules.

Eliz.

 

[Lime]

Ok. I've got it figured out.

But I don't see how it goes from this:

= log_b (3) + log_b (b^2) - log_b (5^3)

to this:

= log_b (3) + 2 log_b (b) - 3 log_b 5

I mean I can see that the exponent 2 goes in front of the second part, and the exponent 3 goes in front of the third part. But what rule allows for this to happen?

 

[stapel]

I don't see how it goes from this:

= log_b (3) + log_b (b^2) - log_b (5^3)

to this:

= log_b (3) + 2 log_b (b) - 3 log_b 5

Try using the "power" rule, as was explained and illustrated in the lesson in the link. (It's the third of the three rules that were mentioned earlier.)

Eliz.