TPRH Passage 14 Gen Chem?? Understanding log

[Padfoot]

On question 6 page 290, I think I have to set logC=KE.
Is it okay to "e" both sides, so it becomes e^logC=e^KE ?

That's the answer, but I checked online and they only added the e to natural log problems.

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[Cawolf]

If you are trying to solve a "log" problem it is standard for log to mean "log base 10".

Where if you want to solve for C in log (C) = KE you would raise both sides on 10.

So 10^(logC) = 10^KE and C = 10^KE

 

[amy_k]

Yes we would only use e if we had ln(C) not log(C), since e^ln cancels out the ln. For log(C) we use 10^ on each side as Cawolf said above^^^, because the 'log' is actually log base 10 and thus 10^[log base 10] cancels out.

 

[Padfoot]

If you are trying to solve a "log" problem it is standard for log to mean "log base 10".

Where if you want to solve for C in log (C) = KE you would raise both sides on 10.

So 10^(logC) = 10^KE and C = 10^KE

The answer was that C = e^KE, so I was confused how they got there. (By = I mean proportional to, if that makes a difference)

 

[amy_k]

What's the original question?

 

[Padfoot]

What's the original question?

The question is: Given that C is the number of detected alpha particles emitted by 228Th and KE is their kinetic energy, then the equation for the dashed line superimposed on the first set of data should be of the form:

The graph has energy(MeV) as the x-axis and "counts per hour" as the y-axis. The y-axis is measured from 10^2, 10^3, all the way to 10^6 .
I assumed that the x-axis represented KE and the y-axis represented logC?

Thanks for your time!