Background Information on
Exponentials and Logarithms
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Since the treatment of the decay of radioactive nuclei is inextricably
linked to the mathematics of exponentials and logarithms, it is important
that you have some expertise in using them. This is intended as a brief
outline of how to use them. Note that you calculator will have these operations
built into it, allowing you to get the result of the operation with a single
key stroke.
We might note in passing that the mathematics below is not all limited
to the topic in the class, that is radioactive decay. It is not even limited
to physics or the other sciences. For example if you establish an account
which pays compound interest at a fixed rate, and reinvest the interest
into the account, then the value of the account increases exponentially
with time, following all of the same mathematical formulae below.
Step 1  Integral Powers of 10
Let us start with something with which (I hope) you are already familiar,
that is powers of 10. The notation 10^{x} means 10 multiplied by
itself x times,

10² = 10 * 10 = 100

10³ = 10 * 10 * 10 = 1000

10^{6} = 10 * 10 * 10 * 10 * 10 * 10 = 1,000,000

and so on
Step 2  NonIntegral Powers of 10
Let us take the previous idea on step further, suppose that the power x
is not a whole number, say 4.518. Then what is 10^{x} now? You
will need your calculator to solve this one. Depending on your calculator
the exact keystrokes will be different, but there is a good chance that
there is a key labeled 10^{x}. Try entering 4.518 followed by 10^{x},
and you should get the result 10^{4.518 }= 32960.97
Step 3  Base 10 Logarithms
The previous step solved the equation y = 10^{x}, if you are given
x then you can calculate y. If x = 4.518 then y = 32960.97. However, suppose
you are given y and are required to calculate x. For example what is the
value of x if y = 69. As an equation solve for the variable x if 69 = 10^{x}.
The key to solving this equation is the logarithm (abbreviated to just
log). It is defined as the opposite operation to the power of 10:
If y = 10^{x} then x = log y
Again your calculator probably has a key labeled log which performs
this operation for you. Type in 69 followed by log, and you should get
the result 1.838849. Just a check try using the method above and you should
get the result 10^{1.838849} = 69. (It might not be exactly 69
because the number 1.838849 has been rounded off a little, but it should
be very close.)
You might want to try the following for practice:

solve for the variable x if 6.92 = 10^{x}. Answer, x = 0.8401

solve for the variable x if 259.9 = 10^{x}. Answer, x = 2.4148

solve for the variable x if 679333 = 10^{x}. Answer, x = 5.8320

solve for the variable x if 45.29 = 10^{x}. Answer, x = 1.6560

solve for the variable x if 7.14 x 10^{12} = 10^{x}. Answer,
x = 12.8537

solve for the variable x if 10000000 = 10^{x}. Answer, x = 7
Step 4  Negative powers
In all the examples above x was a positive number, in which case 10^{x}
is always greater than 1. The value of x could be negative, say x = 0.55.
In that case y = 10^{x} is less than 1. Your calculator can solve
these in the same manner. Type in 0.55, then the change sign key, and then
the 10^{x} key. You should get the result 0.281838.
You might want to try the following for practice:

If x = 0.1 what is y = 10^{x}? Answer, y = 0.79433

If x = 4 what is y = 10^{x}? Answer, y = 0.0001

If x = 2.6754 what is y = 10^{x}? Answer, y = 0.00211

If x = 6.3912578 what is y = 10^{x}? Answer, y = 0.0000004062021

If x = 9.1 what is y = 10^{x}? Answer, y = 7.94 x10^{10}
Negative powers are handled by your calculator also. Suppose we want to
solve 0.01234 = 10^{x}. Rearranging the equation using the definition
of the logarithm from above, this becomes x = log0.01234. Try typing in
0.01234 into your calculator and press the log key. You should get the
answer x = 1.90868
You might want to try the following for practice:

What is x if 0.75 = 10^{x}? Answer, x = log 0.75 = 0.12494

What is x if 0.0000023 = 10^{x}? Answer, x = log 0.0000023 = 5.6383

What is x if 0.012 = 10^{x}? Answer, x = log 0.012 = 1.9208

What is x if 5.4 x 10^{14} = 10^{x}? Answer, x = log 5.4
x 10^{14} = 13.2676
Step 5  Dealing with bases other than 10
In all the above expressions we have used base 10, that is y = 10^{x}.
However there is nothing special about base 10, except that it is the base
that you are used to using, the decimal system. We could have used any
base number such as 2 (the binary system) or 8 (octal system) or 16 hexadecimal
system, or 11, or anything else

2^{5} = 2*2*2*2*2 = 32

3^{6} = 3*3*3*3*3*3 = 729

16^{4} = 16*16*16*16 = 65536
Again these work even if the exponent is not an integer

2^{8.81} = 448.8 (Your calculator should have a y^{x} key
to perform this operation. Type 2, then y^{x}, then 8.81, then
the = key)

16^{9.45} = 239295116727.8 (Type 16, then y^{x}, then 9.45,
then the = key)

3^{0.45} = 0.610 (Type 3, then y^{x}, then 0.45, then
the change sign key, then the = key)
Step 6  Non integral bases
So far we have considered the case when the power is not an integer. Can
we do the same if the base is not an integer? The answer is yes.

4.234³ = 4.234*4.234*4.234 = 75.901884904 (Type 4.234, then y^{x},
then 3, then the = key)

(0.023)^{5} = (0.023)*(0.023)*(0.023)*(0.023)*(0.023) = 0.000000006436343

4.123^{7.76} = 59436.8 (Type 4.123, then y^{x}, then 7.76,
then the = key)
Step 7  e
One of these noninteger numbers is so special in mathematics it is given
its own symbol, e which stands for 2.7182818284590452353602874713527......
(The reasons why it is so special have to do with calculus and need not
concern us here.) However, it in all respects it behaves just like all
other bases

You can raise e to any power. In this case the power is also referred to
as an exponent. Your calculator should have a key to do this for you. It
is likely labeled e^{x} or exp, depending on brand.

e² = 7.3891

e^{7.81} = 2465.13

e^{5.6} = 0.003698

e^{0.004} = 1.004008

You can take logarithms using e as the base. Note, when taking logarithms
above we were using base 10, and the logarithm should more precisely be
written as log_{10} since we could use any base we like. For example
in base 4 we would write log_{4}, and in base 16 we would write
log_{16}. In base e we should write log_{e}, but since
e is special this is given its own designation, ln. Your calculator will
likely also have a key labeled ln also.

if 5 = e^{x} then x = ln(5) = 1.609. (Type 5 and then the ln key.
You should not have to hit the = key to get the answer.)

if 6778.3 = e^{x} then x = ln(6778.3) = 8.8215

if 0.00045 = e^{x} then x = ln(0.00045) = 7.706

if 5.69 x 10^{12} = e^{x} then x = ln(5.69 x 10^{12})
= 29.370

if 4.12 x 10^{31} = e^{x} then x = ln(4.12 x 10^{31})
= 69.964
Step 8  Putting powers and logarithms together
A simple rule is that, for the same base, raising the base to a
power and finding the logarithm are opposite actions. One followed by the
other has no overall effect

log_{y}(y^{x}) = x

log_{10}(10^{4}) = 4

log_{4}(4^{9.56}) = 9.56

log_{e}(e^{12}) = ln(e^{12}) = 12

log_{e}(e^{0.562}) = ln(e^{0.562}) = 0.562

y^{log}_{y}^{(x)} = x

4^{log}_{4}^{(12)} = 12

10^{log(3.456)} = 3.456

e^{ln(2)} = 2

e^{ln(0.00045123)} = 0.00045123
Note though that this rule does not work if the bases are different. For
example ln(10³) is not three. The power operation used base 10, but
the logarithm used base e.
Step 9  A radioactive half life problem
The equation which describes the radioactive decay of an nucleus is N =
N_{o} e^{kt} where N_{o} is the initial number
of nuclei, N is the number remaining after a time t, and k is a constant
related to the lifetime (k = ln2/T = 0.6931/T). With a little bit of rearranging
this can be written as y = e^{x} if y stands for the fraction N/N_{o}
and x stands for kt. Using the rules developed above we can solve these
radioactive decay problems.

Example I An isotope has a half life of 25 minutes. What fraction
remains after one hour?

First of all we need to find k. Its value is ln2/T = ln2/25 = 0.6931/25
= 0.027726

Putting in t = 60 minutes, we have x = kt = 0.027726*60 = 1.66355

Then y = e^{x} = e^{1.66355} = 0.189 = 18.9 %

Example II For this same isotope how much would remain the next
day

Now put t = 24 hours, which is the same as 24*60 = 1440 minutes

Then x = kt = 0.027726*1440 = 39.9253

The fraction remaining is now y = e^{x} = e^{39.9253}
= 4.58 x10^{18}, which is a very small fraction. This sample
has now almost completely disappeared.

Example III Suppose that you have a sample to be dated using ^{14}C.
By measuring the activity of the sample you discover that it has only one
third the activity it must have had when it was new. How old is it?

First of all calculate k. The half life of ^{14}C is 5730 years,
and so k = ln2/T = 0.6931/5730 = 1.20968 x 10^{4}.

In this case we know y = = 0.333333, and we need to solve for x knowing
that 0.333333 = e^{x}.

If y = e^{x} then x = ln y = ln 0.333333 = 1.0986

Since x stands for kt, to calculate t we have t = x/k = (1.0986)/1.20968
x 10^{4} = 9082 years.
As mentioned above this mathematical treatment relates to similar problems
in many walks of life. Let's go through the following examples:

Financial Example 1 You put $1000 into an account which bears an
annual interest rate of 6% paid monthly, and reinvest the interest into
the account. How much is the account worth after 5 years? (Assuming that
there are no subsequent deposits or withdrawals.)

The monthly interest rate is 6%/12 = ½% = 0.005. After one month
the account will have a value of $1000*(1+0.005) = $1000*(1.005), after
2 months $1000*(1.005)*(1.005) = $1000*(1.005)², after 3 months $1000*(1.005)²*(1.005)
= $1000*(1.005)³, and so on. Generalizing this, after x months the
value of the account is $1000*(1.005)^{x}. We have then same mathematics
as above (y = b^{x}) if we use a base of 1.005 for the calculations.

After 5 years (60 months) the value of the account will be $1000*(1.005)^{60}
= $1000*1.34885 = $1348.85

Financial Example II How long will it take for the value of the
account to rise to $1,000,000?

We have to solve the equation 1,000,000 = 1000*1.005^{n}. After
canceling a factor of 1000 from both sides of the equation this becomes
1000 = 1.005^{n}.

The solution to this equation (see above) is n = log_{1.005} 1000.
If your calculator could do calculations in base 1.005 you could solve
this directly, but this is unlikely. Instead we can make use of a useful
relationship for logarithms, that log_{b}x = lnx/lnb.

For our example n = log_{1.005} 1000 = ln1000/ln1.005 = 6.907755/0.000498754
= 1385 months or a little over 115 years.
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