1. ## Evaluating log

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

2. ## Re: Evaluating log

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 $\log_b(X)=a$ then $b^a=X~.$

3. ## Re: Evaluating log

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

4. ## Re: Evaluating log

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 $b^a=X$ then $b^{-a}=\frac{1}{X}~.$

5. ## Re: Evaluating log

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

6. ## Re: Evaluating log

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 $\log(\sqrt{a})=\tfrac{1}{2}\log(a)$.

$\log(20)=\log(10)+\log(2)$

Now contrary to modern trends it seems that here $\log$ means $\log_{10}$ so $\log(10)=1$.

7. ## Re: Evaluating log

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

8. ## Re: Evaluating log

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?

9. ## Re: Evaluating log

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.