Working with Exponents and Logarithms
What is an Exponent?
The exponent of a number says how many times to use the number in a multiplication.
In this example: 23 = 2 × 2 × 2 = 8
(2 is used 3 times in a multiplication to get 8)
What is a Logarithm?
A Logarithm goes the other way.
It asks the question "what exponent produced this?":
And answers it like this:
In that example:
- The Exponent takes 2 and 3 and gives 8 (2, used 3 times, multiplies to 8)
- The Logarithm takes 2 and 8 and gives 3 (2 makes 8 when used 3 times in a multiplication)
A Logarithm says how many of one number to multiply to get another number
So a logarithm actually gives you the exponent as its answer:
Exponents and Logarithms work well together because they "undo" each other (so long as the base "a" is the same):
They are "Inverse Functions"
So doing one, then the other, gets you back to where you started:
- Doing ax, and then the logarithm, gives you x back again:
- Doing the logarithm, then ax , gives you x back again:
It is too bad they are written so differently ... it makes things look strange.
So it may help you to think of ax as "up" and loga(x) as "down":
- going up, then down, returns you back again: down(up(x)) = x , and
- going down, then up, returns you back again: up(down(x)) = x
Anyway, the important thing is that:
The Logarithmic Function can be "undone" by the Exponential Function.
(and vice versa)
Like in this example:
Example, what is x in log3(x) = 5
We can use an exponent (with a base of 3) to "undo" the logarithm:
|We want to "undo" the log so we can get "x ="|
|Use the Exponential Function (on both sides!):|
|And we know that , so:||x = 35|
|Answer:||x = 243|
Example: Calculate y in y=log4(1/4)
|Use the Exponential Function on both sides:|
|Simplify:||4y = 1/4|
Now a simple trick: 1/4 = 4-1
|So:||4y = 4-1|
|And so:||y = -1|
Properties of Logarithms
One of the powerful things about Logarithms is that they can turn multiply into add.
loga( m × n ) = logam + logan
"the log of a multiplication is the sum of the logs"
Why is that true?
To show you why, we get to use and again:
|First, make m and n into "exponents of logarithms":|
Then use one of the Laws of Exponents
Finally undo the exponents.
It is one of those clever things we do in mathematics which can be described as "we can't do it here, so let's go over there, then do it, then come back"
Have a look over that again so you understand why it happens. But the important thing is that it does work, and it is the result that matters.
Table of Properties
OK, we just used one of the Laws of Exponents that said xm × xn = xm+n, and we can use other Laws of Exponents as well!
Here are some of the most useful results for logarithms:
|loga(mn) = logam + logan||the log of a multiplication is the sum of the logs (as shown above)|
|loga(m/n) = logam - logan||the log of a division is the difference of the logs|
|loga(1/n) = -logan||this just follows on from the previous "division" rule, because loga(1) = 0|
|loga(mr) = r ( logam )||the log of m with an exponent r is r times the log of m|
Remember: the base "a" is always the same!
History: Logarithms were very useful before calculators were invented ... for example, instead of multiplying two large numbers, by using logarithms you could turn it into addition (much easier!)
And there were books full of Logarithm tables to help.
Let us have some fun using them:
Example: Simplify loga( (x2+1)4√x )
|Start with:||loga( (x2+1)4√x )|
|Use loga(mn) = logam + logan :||loga( (x2+1)4 ) + loga( √x )|
|Use loga(mr) = r ( logam ) :||4 loga(x2+1) + loga( √x )|
|Also √x = x½ :||4 loga(x2+1) + loga( x½ )|
|Use loga(mr) = r ( logam ) again:||4 loga(x2+1) + ½ loga(x)|
That is as far as I can simplify it. I can't do anything with loga(x2+1).
Answer: 4 loga(x2+1) + ½ loga(x)
Note: there is no rule for handling loga(m+n) or loga(m-n)
We can also apply the logarithm rules "backwards" to combine logarithms:
Example: Turn this into one logarithm: loga(5) + loga(x) - loga(2)
|Start with:||loga(5) + loga(x) - loga(2)|
|Use loga(mn) = logam + logan :||loga(5x) - loga(2)|
|Use loga(m/n) = logam - logan :||loga(5x/2)|
The Natural Logarithm and Natural Exponential Functions
When the base is e ("Euler's Number" = 2.718281828459...) you get:
- The Natural Logarithm loge(x), and is more commonly written ln(x)
- The Natural Exponential Function ex
And the same idea that one can "undo" the other is still true:
ln(ex) = x
e(ln x) = x
And here are their graphs:
Natural Exponential Function
Graph of f(x) = ln(x)
Graph of f(x) = ex
Passes through (1,0) and (e,1)
Passes through (0,1) and (1,e)
They are the same curve with x-axis and y-axis flipped. Which is another thing to show you they are inverse functions.
On a calculator you can use the "ln" button.
Always try to use Natural Logarithms and the Natural Exponential Function whenever possible.
The Common Logarithm
When the base is 10 you get:
- The Common Logarithm log10(x), which is sometimes written as log(x)
Engineers love to use it, but It is not much used in mathematics.
On a calculator it is the "log" button.
It is handy because it tells you how "big" the number is in decimal ... how many times you need to use 10 in a multiplication.
Example: Calculate log10 100
Well, 10 × 10 = 100, so when 10 is used 2 times in a multiplication you get 100:
log10 100 = 2
Likewise log10 1,000 = 3, log10 10,000 = 4, and so on.
Example: Calculate log10 369
OK, best to use my calculator's "log" button:
log10 369 = 2.567...
Changing the Base
What if you want to change the base of a logarithm?
Easy! Just use this formula:
"x goes up, a goes down"
Or another way to think of it is that logb a is like a "conversion factor" (same formula as above):
loga x = logb x / logb a
So now you can convert from any base to any other base.
Another useful property is:
loga x = 1 / logx a
Did you see how "x" and "a" swap positions?
Example: Calculate 1 / log8 2
1 / log8 2 = log2 8
And 2 × 2 × 2 = 8, so when 2 is used 3 times in a multiplication you get 8:
1 / log8 2 = log2 8 = 3
But you will more often use the Natural Logarithm, so this is worth remembering:
loga x = ln x / ln a
Example: Calculate log4 22
My calculator doesn't have a "log4" button ...
... but it does have an "ln" button, so we can use that:
log4 22 = ln 22 / ln 4 = 3.09.../1.39... = 2.23 (to 2 decimal places)
What does this answer mean? It means that 4 with an exponent of 2.23 equals 22. So we can check that answer:
Check: 42.23 = 22.01 (close enough!)
Here is another example:
Example: Calculate log5 125
log5 125 = ln 125 / ln 5 = 4.83.../1.61... = 3 (exactly)
I happen to know that 5 × 5 × 5 = 125, (5 is used 3 times to get 125), so I expected an answer of 3, and it worked!
Real World Usage
Logarithms are used a lot in the real world. You may have heard of these:
The magnitude of an earthquake is a Logarithmic scale.
The famous "Richter Scale" uses this formula:
M = log10 A + B
Where A is the amplitude (in mm) measured by the Seismograph
and B is a distance correction factor
Nowadays there are more complicated formulas, but they still use a logarithmic scale
Loudness is measured in Decibels (dB for short);
Loudness in dB = 10 log10 (p × 1012)
where p is the sound pressure.
Acidic or Alkaline
Acidity (or Alkalinity) is measured in pH:
pH = -log10 [H+]
where H+ is the molar concentration of dissolved hydrogen ions.
Note: in chemistry [ ] means molar concentration (moles per liter).
Example: Solve 2 log8 x = log8 16
|Start with:||2 log8 x = log8 16|
|Bring the "2" into the log:||log8 x2 = log8 16|
|Remove the logs (they are same base):||x2 = 16|
|Solve:||x = -4 or +4|
But... but ... but ... you cannot have a log of a negative number!
So the -4 case is not defined.
Check: use your calculator to see if this is the right answer ... also try the "-4" case.
Example: Solve e-w = e2w+6
|Start with:||e-w = e2w+6|
|Apply ln to both sides:||ln(e-w) = ln(e2w+6)|
|And ln(ew)=w:||-w = 2w+6|
|Simplify:||-3w = 6|
|Solve:||w = 6/-3 = -2|
Answer: w = -2
Check: e-(-2)= e2 and e2(-2)+6=e2