What am I doing wrong in solving 28/68=b^14 for b?

[goosey00]

The 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?

 

[Jameson]

The 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?

Hi goosey. First you simplify 28/68 which is about .41176. Then you must take that to the (1/14)th power, not to the 14th power. That's because opposites cancel each other out in a way.

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.

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

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

 

[goosey00]

got it now. Ok, so the last part of this question my professor said do not find the solution as the book says by graphing. The final part of the problem is take f(t)=22+68(.93859)^t. He said to do it the way we learned through exponential and the answer is 21. Again, I must be off since I am not getting it right. The whole problem is taking Newtons law to find different temps. at different room temperatures.

 

[Jameson]

got it now. Ok, so the last part of this question my professor said do not find the solution as the book says by graphing. The final part of the problem is take f(t)=22+68(.93859)^t. He said to do it the way we learned through exponential and the answer is 21. Again, I must be off since I am not getting it right. The whole problem is taking Newtons law to find different temps. at different room temperatures.

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.

 

[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?

 

[Sudharaka]

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?

Hi goosey00, :)

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,

93 to 60 after 13 min in a room of 25 degrees.

And how did you obtain, \(.49295^\frac{1}{13}\) ?

 

[goosey00]

Before I get even more confusing, Ill just post the last part on my teachers blog. Thanks for the response. I really appreciate it.

 

[Ackbach]

Another way to solve the original problem is to take the logarithm of both sides and use the properties of logarithms to simplify.