# LOG and EXP Calculations

• Oct 14th 2009, 11:32 AM
darylt
LOG and EXP Calculations
Hi all, I am on a Paramedic Science Degree Course, but I am expected to be able to do a level of maths above my original school training.

My problem is that we are using logarithms and exponential calculations, but until our session this week I had no idea that they even existed...

I am allowed to use a calculator to complete these calculations and would appreciate being told how to do these on the calculator as well as how to do it with the good old pen and paper (if its not too hard).

I have to complete a table of results where the first line is completed for me;

Result 1 is
Concentration of fluid is
.01(1x10-8)

**10-8 the minus 8 is supposed to be "to the power of eight"**

The Log of the solution is
$\displaystyle -8$

I understand that $\displaystyle 10-8$ is the equivalent 8 zeros, but thats about all I understand.

• Oct 14th 2009, 11:39 AM
e^(i*pi)
Quote:

Originally Posted by darylt
Hi all, I am on a Paramedic Science Degree Course, but I am expected to be able to do a level of maths above my original school training.

My problem is that we are using logarithms and exponential calculations, but until our session this week I had no idea that they even existed...

I am allowed to use a calculator to complete these calculations and would appreciate being told how to do these on the calculator as well as how to do it with the good old pen and paper (if its not too hard).

I have to complete a table of results where the first line is completed for me;

Result 1 is
Concentration of fluid is
.01(1x10-8)

**10-8 the minus 8 is supposed to be "to the power of eight"**

The Log of the solution is
$\displaystyle -8$

I understand that $\displaystyle 10-8$ is the equivalent 8 zeros, but thats about all I understand.

The definition of a log is $\displaystyle log_c(b) = a$ means $\displaystyle c^a = b$ as long as c is not 1 or 0 and b > 0.

In the case of $\displaystyle \log_{10}(10^{-8})$ we are trying to find $\displaystyle a$ and we know $\displaystyle c = 10$ and $\displaystyle b = 10^{-8}$

We can use the law of logs that says $\displaystyle log_b(a^k) = k\,log_b(a)$ to bring the -8 to the front of the log.

By definition (as above) $\displaystyle log_b(b) = 1$ so the answer is simply -8

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If you're trying to find the log of $\displaystyle 0.01 \times 10^{-8}$ then the answer won't be -8 because you need to take the 0.01 into account.

As [tex]0.01 = 10^{-2}[/maTH] then $\displaystyle 0.01 \times 10^{-8} = 10^{-10}$ (this is because we can add exponents of the same base when multiplying). If we take the log of the new answer we get -10