# Power rule

• Sep 17th 2011, 01:35 PM
benny92000
Power rule
If a and b are nonzero real numbers and (3.92)a = (7.86)b, what is the value of (b/a)^2? The answer is .44. The answer says use logs, so I took the log of both sides. Then they say to use the power rule. I thought the a and b would have to be an exponent to use the power rule? Do they, or is the answer explanation I'm reading correct?
• Sep 17th 2011, 01:40 PM
mr fantastic
Re: Power rule
Quote:

Originally Posted by benny92000
If a and b are nonzero real numbers and (3.92)a = (7.86)b, what is the value of (b/a)^2? The answer is .44. The answer says use logs, so I took the log of both sides. Then they say to use the power rule. I thought the a and b would have to be an exponent to use the power rule? Do they, or is the answer explanation I'm reading correct?

(3.92)a = (7.86)b

=> b/a = 3.92/7.86

=> (b/a)^2 = 0.2487.

I hate to break it to you sport, but the given answer for the question as posted is wrong (and it does not require logs).

Perhaps you need to go back and post the correct question ....?
• Sep 17th 2011, 01:51 PM
benny92000
Re: Power rule
I think that answers the question. It is a typo (from the part of the book that I'm using). I assume a and b should be exponents.
• Sep 18th 2011, 03:56 PM
HallsofIvy
Re: Power rule
Is there a reason why you chose NOT to tell us that the a and b were exponents to start with? That changes the entire problem.

If $(3.92)^a = (7.86)^b$, then, yes, use logarithms. log(3.92^a)= alog(3.92)= log(7.86^b)= blog(7.86) so that
$\frac{b}{a}= \frac{log(3.92)}{log(7.86)}$
• Sep 18th 2011, 04:17 PM
benny92000
Re: Power rule
It was a typo in the question. Thanks though for the help.