Working With Exponents And Logarithms - Math Is Fun

Working with Exponents and Logarithms

What's an Exponent?

2 with exponent 3

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's a Logarithm?

A Logarithm goes the other way.

It asks the question "what exponent produced this?":

Logarithm Question

And answers it like this:

exponent to logarithm

In that example:

  • The Exponent takes 2 and 3 and gives 8 (2, used 3 times in a multiplication, makes 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 us the exponent as its answer:

Logarithm base 2 of 8 equals 3, where 3 is the exponent

(Also see how Exponents, Roots and Logarithms are related.)

Working Together

Exponents and Logarithms work well together because they "undo" each other (so long as the base "a" is the same):

Flowchart showing x goes to a^x and log base a of x returns to x

They are "Inverse Functions"

Doing one, then the other, gets us back to where we started:

Doing ax then loga gives us back x:loga(ax) = x Doing loga then ax gives us back x:aloga(x) = x

It is too bad they are written so differently ... it makes things look strange. So it may help to think of ax as "up" and loga(x) as "down":

going up then down returns us back again:down(up(x)) = x going down then up returns us back again:up(down(x)) = x

Anyway, the important thing is that:

The Logarithmic Function is "undone" by the Exponential Function.

(and vice versa)

Like in this example:

Example, what's x in log3(x) = 5

We want to "undo" the log3 so we can get "x ="

Start with: log3(x) = 5 Use the Exponential Function on both sides:3log3(x) = 35 And we know that 3log3(x) = x, so:x = 35 Answer: x = 243

And also:

Example: Calculate y in y = log4(14)

Start with:y = log4(14) Use the Exponential Function on both sides:4y = 4log4(14) Simplify:4y = 14 Now a simple trick: 14 = 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 multiplication is the sum of the logs"

Why is that true? See Footnote.

Using that property and the Laws of Exponents we get these useful properties:

the log of multiply is the sum of the logs: loga(m × n) = logam + logan the log of divide is the difference of the logs: loga(m/n) = logam − logan that rule also gets us the reciprocal rule: loga(1/n) = −logan as loga(1) = 0 the log of m with an exponent r is r times the log of m: loga(mr) = r ( logam )

Remember: here the base "a" is always the same!

book of logarithmsHistory: Logarithms were very useful before calculators were invented ... for example, instead of multiplying two large numbers, by using logarithms we can turn it into addition (much easier!)

And there were books full of Logarithm tables to help.

Let's have some fun using the properties:

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's as far as we can simplify it ... we can't do anything with loga(x2+1)

Answer: 4 loga(x2+1) + ½ loga(x)

Note: there's 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)

Answer: loga(5x/2)

The Natural Logarithm and Natural Exponential Functions

When the base is Euler's Number e = 2.718281828459... we get:

  • The Natural Logarithm loge(x) which is more commonly written ln(x)
  • The Natural Exponential Function ex

And, yes, one can "undo" the other:

ln(ex) = x

e(ln x) = x

And here are their graphs:

Natural Logarithmnatural logarithm function Graph of f(x) = ln(x)

Passes through (1,0) and (e,1)

Natural Exponential Functionnatural exponential function Graph of f(x) = ex

Passes through (0,1) and (1,e)

They are the same curve with x-axis and y-axis flipped.

See them together here: images/function-graph.js?fn0=e^(x)&fn1=ln(x)&xmin=-5.294&xmax=5.259&ymin=-1.447&ymax=4.553&vara=1|0|10&vark=1|0|2

Which also shows us they are inverse functions.

calculator ln button

On a calculator the Natural Logarithm is the "ln" button.

Always try to use Natural Logarithms and the Natural Exponential Function whenever possible.

The Common Logarithm

When the base is 10 we get:

  • The Common Logarithm log10(x), which is sometimes written as log(x)
calculator log button

On a calculator the Common Logarithm is the "log" button.

Engineers love to use it, but it isn't used much in mathematics.

The Common Logarithm is handy because it tells us how "big" the number is in decimal (how many times we 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 we 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 we want to change the base of a logarithm?

We can use this formula:

Log Change Base

"x goes up, a goes down"

1logb a works as a "conversion factor" from one base to any other base.

Another useful property is:

loga x = 1 / logx a

Log Reciprocal

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 we get 8:

1 / log8 2 = log2 8 = 3

And the Natural Logarithm gives us this simple Change of Base version:

loga x = ln xln a

Example: Calculate log4 22

calculator ln button

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's another example:

Example: Calculate log5 125

We can use the "ln" function on the calculator:

log5 125= ln 125 / ln 5 = 4.83... / 1.61... = 3.00 (to 2 decimal places)

Is it exactly 3? We should not trust a calculator as there could be rounding errors, but in this case we can check that 53 = 5 × 5 × 5 = 125 exactly, so:

Answer: 3

Real World Usage

Here are some uses for Logarithms in the real world:

Earthquakes

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.

Sound

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).

More Examples

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 ... we can't take the log of a negative number!

So the −4 case isn't defined.

Answer: 4

Check: use a calculator to see if this is the right answer ... also try the "−4" case.

Example: Solve ew = 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

Footnote: Why does log(m × n) = log(m) + log(n) ?

To see why, we'll use aloga(x) = x and loga(ax) = x like this:

Log Product Rule

So we seem to make things more complicated by transforming into aloga(x) but then we are able to add them, then we transform back again and we have a solution!

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, do it, then come back".

585, 1234, 587, 1237, 8137, 8221, 8243, 8244, 8138, 8222 Exponents Logarithms Algebra 2 Index

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