Thread: Evaluating log

If log_b 6 = .4040 then b^-.4040 is equal to...I am not to sure about how to solve this.

Originally Posted by Bashyboy

If log_b 6 = .4040 then b^-.4040 is equal to...I am not to sure about how to solve this.

If then

Well, the exponent on b is negative, does that change the definition you gave me at all?

Originally Posted by Bashyboy

Well, the exponent on b is negative, does that change the definition you gave me at all?

No indeed. It just requires one more step.

If then

Oh, yes--I undserstand now. I do have another one, though: If log 2 = .3010 then log sqroot(20) is equal to.

Originally Posted by Bashyboy

Oh, yes--I undserstand now. I do have another one, though: If log 2 = .3010 then log sqroot(20) is equal to.

First, you should start a new thread for a new question.

Note that .



Now contrary to modern trends it seems that here means so .

Do I sense condescension in your writing, or am I thoroughly mistaken?

Originally Posted by Bashyboy

Do I sense condescension in your writing, or am I thoroughly mistaken?

You either have the largest chip on your shoulder or you can't read.
Are you angry at me for not giving a complete and polished solution?
Do you not want to learn how to do these?

No, I am absolutely not angry. I inferred that impression on my first reading of your post; but, after re-reading it, I have found that impression to be wrong. I am terrible sorry. I sensitive when it comes to scorning my mathematical knowledge; and I know it is scarce, due to myself having studied it a little later in life than normal, and that is why I was a bit jumpy. Again, sorry.