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- Jan 26, 2012

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Hi goosey. First you simplify 28/68 which is about .41176. Then you must take that to theThe problem is 28/68=b^14 and I need to solve for b. I get .41176^14 which is not the right answer(which is .93859) What am I doing wrong?

If you have something squared and you take the square-root of it, the two things cancel and you just get the original number. \(\displaystyle \sqrt{5^2}=5\)

When working with higher powers, we you can use the reciprocal exponents to cancel out things. For the tenth power, \(\displaystyle \left( 4^{10} \right) ^\frac{1}{10} = 4\)

In our problem we have \(\displaystyle .41176=b^{14}\) and we want to find "b". So if we take both sides to the (1/14) power then the right side will simplify nicely.

\(\displaystyle .41176^{\frac{1}{14}}=\left( b^{14} \right) ^\frac{1}{14}\)

\(\displaystyle .41176^{\frac{1}{14}}=b\)

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f(t) represents a temperature after time "t", but we don't know what temperature we're looking for. We need to have a temperature before we can find at what time it reaches that temperature.

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Hi goosey00,93 to 60 after 13 min in a room of 25 degrees. how long before it gets to 40 degrees. Although this is a different problem I am stuck at the same point. so far I have .49295 (1/13) which I got .94704. Now what is it do I do now w/out graphing to get it?

Can you please clarify a bit more goosey. You have the equation, \(f(t)=22+68(.93859)^t\) where \(f(t)\) represents the temperature and \(t\) the time required to reach that temperature. What do you mean by,

And how did you obtain, \(.49295^\frac{1}{13}\) ?93 to 60 after 13 min in a room of 25 degrees.

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