Lesson LOGARITHMS AND EXPONENTIAL AND LOGARITHMIC EQUATIONS

Source code of 'LOGARITHMS AND EXPONENTIAL AND LOGARITHMIC EQUATIONS'

This Lesson (LOGARITHMS AND EXPONENTIAL AND LOGARITHMIC EQUATIONS) was created by by Theo(13342)

This lesson covers an overview of LOGARITHMS AND EXPONENTIAL AND LOGARITHMIC EQUATIONS

REFERENCES

<a href = "http://www.themathpage.com/aPreCalc/logarithms.htm#laws" target = "_blank">http://www.themathpage.com/aPreCalc/logarithms.htm#laws</a>
<a href = "http://www.sosmath.com/algebra/logs/log1/log1.html" target = "_blank">http://www.sosmath.com/algebra/logs/log1/log1.html</a>
<a href = "http://www.themathpage.com/aPreCalc/logarithms.htm" target = "_blank">http://www.themathpage.com/aPreCalc/logarithms.htm</a>
<a href = "http://www.purplemath.com/modules/logs.htm" target = "_blank">http://www.purplemath.com/modules/logs.htm</a>
<a href = "http://people.hofstra.edu/Stefan_Waner/Realworld/calctopic1/logs.html" target = "_blank">http://people.hofstra.edu/Stefan_Waner/Realworld/calctopic1/logs.html</a>
<a href = "http://www.themathpage.com/aPreCalc/logarithmic-exponential-functions.htm" target = "_blank">http://www.themathpage.com/aPreCalc/logarithmic-exponential-functions.htm</a>
<a href = "http://www.purplemath.com/modules/solvexpo2.htm" target = "_blank">http://www.purplemath.com/modules/solvexpo2.htm</a>
<a href = "http://www.occc.edu/maustin/exp_log_equations/exponential%20and%20logarithmic%20equations.htm" target = "_blank">http://www.occc.edu/maustin/exp_log_equations/exponential%20and%20logarithmic%20equations.htm</a>
<a href = "http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut46_logeq.htm" target = "_blank">http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut46_logeq.htm</a>
<a href = "http://tutorial.math.lamar.edu/Classes/Alg/SolveLogEqns.aspx" target = "_blank">http://tutorial.math.lamar.edu/Classes/Alg/SolveLogEqns.aspx</a>
<a href = "http://www.purplemath.com/modules/solvexpo.htm" target = "_blank">http://www.purplemath.com/modules/solvexpo.htm</a>
<a href = "http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut45_expeq.htm" target = "_blank">http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut45_expeq.htm</a>

Logarithmic Equations and Exponential Equations are two sides of the same coin.  Logarithms are used to solve exponential equations and exponents are used to solve logarithmic equations.

For more information on exponents, see the lesson on EXPONENTS.

BASIC DEFINITION OF LOGARITHMS

{{{log(b,x) = y}}} if and only if {{{b^y = x}}}

This means that the log of x to the base b equals y if and only if the base b raised to the power of the exponent y = x.

Example:
let b = 2 and x = 8 and y = 3

{{{log(2,8) = 3}}} if and only if {{{2^3 = 8}}}

This means that the log of 8 to the base 2 = 3 if and only if 2 raised to the third power = 8.

Since {{{2^3 = 8}}}, this is confirmed to be true.

RULES OF LOGARITHMS

There are three major rules of logarithms.

LOGARITHM RULE NUMBER 1

{{{log(b,(x*y)) = log(b,(x)) + log(b,(y))}}}

This means that the log of (x*y) to the base b equals the log of (x) to the base b plus the log of (y) to the base b.

Example:
{{{log(2,(8*16)) = log(2,(8)) + log(2,(16))}}}

This means that the log of (8*16) to the base 2 = log of 8 to the base 2 plus the log of 16 to the base 2.

{{{log(2,(8*16))}}} = {{{log(2,(128)) = 7}}} because {{{2^7 = 128}}}.
{{{log(2,(8)) = 3}}} because {{{2^3 = 8}}}.
{{{log(2,(16)) = 4}}} because {{{2^4 = 16}}}.
{{{log(2,(8)) + log(2,(16))}}} = {{{(3+4) = 7}}} = {{{log(2,(8*16))}}} confirming the rule.

LOGARITHM RULE NUMBER 2

{{{log(b,(x/y)) = log(b,(x)) - log(b,(y))}}}

This means that the log of (x/y) to the base b equals the log of (x) to the base b minus the log of (y) to the base b.

Example:
{{{log(2,(16/8)) = log(2,(16)) - log(2,(8))}}}

This means that the log of (16/8) to the base 2 = log of 16 to the base 2 minus the log of 8 to the base 2.

{{{log(2,(16/8))}}} = {{{log(2,(2)) = 1}}} because {{{2^1 = 2}}}.
{{{log(2,(16)) = 4}}} because {{{2^4 = 16}}}.
{{{log(2,(8)) = 3}}} because {{{2^3 = 8}}}.
{{{log(2,(16))}}} - {{{log(2,(8))}}} = {{{(4-3) = 1}}} = {{{log(2,(16/8))}}} confirming the rule.

LOGARITHM RULE NUMBER 3

{{{log(b,(x^n)) = n*log(b,(x))}}}

This means that the log of (x^n) to the base b equals n times the log of (x) to the base b.

Example:
{{{log(2,(8^5)) = 5*log(8)}}}

This means that the log of (8^5) to the base 2 = 5 times the log of 8 to the base 2.

{{{log(2,(8^5))}}} = {{{log(2,(32768)) = 15}}} because {{{2^15 = 32768}}}.
{{{log(2,(8)) = 3}}} because {{{2^3 = 8}}}.
{{{5*log(2,(8))}}} = {{{5*3 = 15}}} = {{{log(2,(8^5))}}} confirming the rule.

USING YOUR CALCULATOR TO SOLVE LOGARITHMS

Since most commonly sold scientific calculators can solve common logarithms (base 10) or natural logarithms (base e), it make sense to use these bases as often as possible.

If you are given a problem in logarithms taken to a base other than 10 or e, you can convert that logarithm to a base 10 or a base e by using the conversion formula to be discussed next.

CHANGING THE BASE OF A LOGARITHM

You can change the base of a logarithm from any base to any other base using the following conversion formula:

{{{log(b,(x)) = log(c,(x))/log(c,(b))}}}

This means that the log of x to the base b equals the log of x to the base c divided by the log of b to the base c.

Example:

Let b = 2 and let c = 10 and let x = 8.

{{{log(2,(8)) = log(10,(8))/log(10,(2))}}}

We know that {{{log(2,8))}}} = 3 because {{{2^3}}} = 8 so we didn't really need to convert in this case but it serves to demonstrate that the conversion formula works as shown below:

Converting to the base 10 allows us to solve this directly using the LOG function of the calculator.

{{{log(10,(8)) = .903089987}}}
{{{log(10,(2)) = .301029996}}}

{{{log(10,(8))/log(10,(2))}}} = {{{.903089987/.301029996 = 3}}}

since {{{log(2,(8)))}}} = 3 and {{{log(10,(8))/log(10,(2))}}} = 3 the two equations are equivalent confirming that the formula for changing the base of a logarithm is valid.

For more information on changing the base of a logarithm, see the lesson on CHANGING THE BASE OF A LOGARITHM.

BASIC RULES OF ALGEBRAIC AND ARITHMETIC OPERATIONS AS THEY APPLY TO LOGARITHMS

If a = b, then {{{log((a)) = log((b))}}}

This means that if a = b, then log of a = log of b.

Example:
let a = 5 and b = 5

If 5 = 5, then log(5) = log(5)

Taking the log of both sides of an equality does not change the equality.

This fact alone is used to solve many exponential equations.

SOLVING PROBLEMS USING LOGARITHMS

Logarithms are commonly used to solve problems where the unknown value is the exponent.  These would be called exponential equations.

EXAMPLE NUMBER 1

{{{e^x = 5000}}}

It appears that the easiest way to solve this is to use the basic definition of logarithms that states the following:

{{{e^x = 5000}}} is true if and only if {{{log(e,(5000)) = x}}}

Since {{{log(e,(5000))}}} is the same as {{{ln(5000)}}}, we can solve this directly using the LN function of our scientific calculator.

Our answer is {{{ln(5000) = 8.517193191}}}.

This answer is confirmed because {{{e^(8.517193191) = 5000}}}.

The other way to solve this would have been to take the log of both sides of the equation.  Natural log appears to be the best way since this equation is already in base e format.

The equation of {{{e^x = 5000}}} becomes {{{log(e,(e^x)) = log(e,(5000))}}} which becomes {{{ln(e^x) = ln(5000)}}}.

This is because natural log (ln) of a number means the same thing as the log to the base e of a number.

We would use the LN function of the scientific calculator.

By rule number 3, {{{ln(e^x)}}} is the same as {{{x*ln(e)}}} so our equation becomes {{{x*ln(e) = ln(5000)}}}.

Since {{{ln(e) = 1}}} and {{{ln(5000) = 8.517193191}}}, our equation becomes {{{x = 8.517193191}}} and we need go no further.

Our answer is {{{x = 8.517193191}}}

This answer is confirmed because {{{e^(8.517193191) = 5000}}}.

EXAMPLE NUMBER 2

You are given the equation of {{{2^x = 5^y}}} and told to solve for y.

It appears that the easiest way to solve this is to take the common or natural log of both sides of the equation and solve using the LOG function of the calculator or the LN function of the calculator.

We'll use the LOG function of the scientific calculator this time.

Our equation of {{{2^x = 5^y}}} becomes {{{log(10,(2^x)) = log(10,(5^y))}}} which becomes {{{log((2^x)) = log((5^y))}}}.

This is because the log of a number means the same thing as the log to the base 10 of a number.
If the base of a log is not mentioned, it is assumed to be 10.

by rule number 3, this equation becomes {{{x*log((2)) = y*log((5))}}}

Solving for y, we get {{{y = (x*log((2)))/(log((5)))}}} which becomes {{{y = ((.301029996)/(.698970004))*x)}}} which becomes {{{y = (.430676558)*(x)}}}

Solving for x, we get {{{x = (y*log((5)))/(log((2)))}}} which becomes {{{x = ((.698970004)/(.301029996))}}} which becomes {{{x = (2.321928095)*(y)}}}

What this says is that if we know the value of x we can find the value of y and, conversely, if we know the value of y we can find the value of x.

assume y = 1
if {{{y = 1}}}, then {{{x = 2.321928095}}}
{{{2^x = 5^y}}} becomes {{{2^2.321928095}}} = {{{5^1}}} = 5 confirming that the ratio between y and x is correct.

assume x = 1
if {{{x = 1}}}, then {{{y = .430676558}}}
{{{2^x = 5^y}}} becomes {{{2^1}}} = {{{5^.430676558}}} = 2 confirming that the ratio between x and y is correct.

EXAMPLE NUMBER 3

You are given {{{log(3,(3486784401)) = y}}} and asked to solve for y.

You can't use the LOG function of the calculator because the base is not 10.
You can't use the LN function of the calculator because the base is not e.

It appears that the easiest way to solve this is to use the CHANGE OF BASE formula.

That formula is {{{log(b,(x)) = log(c,(x))/log(c,(b))}}}

You have:
b = 3
x = 3486784401
You let:
c = 10 because that can be solved using the LOG function of your calculator.

Your equation of {{{log(3,(3486784401))}}} becomes {{{log(10,(3486784401))/log(10,(3))}}} which becomes {{{log((3486784401))/log((3))}}} because the log without the base indicated means the log to the base 10.

{{{log((3486784401))/log((3))}}} = {{{9.542425094 / .477121255}}} = {{{20}}}.

Your answer is 20.
This is confirmed because {{{3^20 = 3486784401}}}

The other way to solve this would have been to convert it to exponential form and then take the common log of both sides of the equation or the natural log of both sides of the equation.

Your original equation of {{{log(3,(3486784401)) = y}}} is true if and only if {{{3^y = 3486784401}}} by the basic definition of logarithms.

We'll take the natural log of both sides of {{{3^y = 3486784401}}} this time.  
Either common or natural logs will work.

{{{3^y = 3486784401}}} becomes {{{log(3,(3^y)) = log(3,(3486784401))}}} which becomes {{{ln((3^y)) = ln((3486784401))}}} because natural log (ln) of a number means the same thing as taking the log of that number to the base e.

By rule number 3, {{{ln((3^y)) = y*ln((3))}}} so our equation becomes {{{y*(ln((3))) = ln((3486784401))}}}

Using our calculator, {{{y*(ln((3))) = ln((3486784401))}}} becomes {{{y*1.098612289 = 21.97224577}}}.

Solving for y, this equation becomes {{{y = 21.97224577/1.098612289}}} which becomes {{{y = 20}}}.

Our answer is y = 20

This is confirmed because {{{3^20 = 3486784401}}}

SOME ADDITIONAL EXAMPLES

Most of the following examples are from the wtamu website mentioned as one of the major references for this lesson.

One example is from their section on exponential equations and some more examples are from their section on logarithmic equations.

The difference between exponential equations and logarithmic equations is as follows:

Exponential equations are equations where the unknown value is in the exponent.
Logarithmic equations are equations where the unknown value is in the logarithm.
The difference will be apparent as we go through the next series of examples.

EXAMPLE NUMBER 4

Given the equation {{{e^2x - 2e^x - 8 = 0}}} solve for x

The trick according to the folks at wtamu is to isolate the exponential expression on one side of the equation and then solve using log functions of your calculator.  LN or LOG will do, but they used LN in their example so I will continue along those lines.

We can factor the left side of this equation to get {{{(e^x-4)*(e^x+2) = 0}}}
This makes {{{e^x = 4)}}} or {{{e^x = -2}}}
We can use the LN function of our calculator to solve each of these in turn.  We'll look at {{{e^x = 4)}}} first.
We do that by taking the natural log of both sides of this equation to get:

{{{ln((e^x)) = ln((4))}}}

Since {{{ln((e^x))}}} becomes {{{x*ln((e))}}} by rule number 3, our equation becomes {{{x*ln((e)) = ln((4))}}} which becomes {{{x = ln((4))}}} because {{{ln((e)) = 1}}}.

One answer for x is {{{ln((4)) = 1.386294361}}}

This is confirmed because {{{e^(1.386294361) = 4}}}

We now look at {{{e^x = -2}}}
Since anything to a positive base has to be greater than 0, this answer is invalid.
If we didn't recognize this, we would have tried to take the natural log of both sides of this equation to get:
{{{ln((e^x)) = ln((-2))}}}
As soon as we tried to get {{{ln((-2))}}} we would have gotten an error because you can't take the log of any number that is negative or 0.  It has to be positive.

The only good answer here is therefore {{{x = 1.386294361}}}

Up to now we've been solving exponential equations using logarithms.
Now we'll solve logarithmic equations using exponents.
The difference is in where the unknown value is.  If the unknown value is in the exponent, it's an exponential equation.  If the unknown value is in the logarithm, it's a logarithmic equation.

EXAMPLE NUMBER 5

Given the equation {{{log(5,(x+2)) = 3}}} solve for x

The trick according to the folks at wtamu is to isolate the logarithmic expression on one side of the equation and then solve using exponential functions of your calculator.

Note that the unknown value is in the logarithmic form of the equation and not in the exponential form of the equation as you will see shortly.

Since the logarithmic expression is already isolated in this example, we can convert this to the exponential form by using the basic definition of logarithms.

Our equation of {{{log(5,(x+2)) = 3}}} is true if and only if {{{5^3 = (x+2)}}}
Since {{{5^3}}} = 125 then our equation becomes {{{125 = x+2}}} which becomes {{{x = 123}}}

Our answer is x = 123

This is confirmed because {{{log(5,(123+2))}}} = {{{log(5,(125))}}} = {{{3}}} which is true because {{{5^3 = 125}}}.

EXAMPLE NUMBER 6

Given {{{log((x+21)) + log((x)) = 2}}} then solve for x

We'll be dealing with common logs here because {{{log((number))}}} means {{{log(10,((number)))}}}.

In this equation, rule number 1 of logarithms will be used.

Rule number 1 states that {{{log(b,(x*y)) = log(b,(x)) + log(b,(y))}}}

Our equation of {{{log((x+21)) + log((x)) = 2}}} becomes {{{log(((x+21)*(x))) = 2}}} which becomes {{{log((x^2 + 21x)) = 2}}}

By the basic definition of logarithms, this can only be true if {{{10^2 = x^2 + 21x}}}

Since {{{10^2 = 100}}} we can subtract 100 from both sides of this equation to get {{{x^2 + 21x - 100 = 0}}}

This factors out to be {{{(x+25)*(x-4) = 0}}} which makes x = 4 or x = (-25).

If x = 4, then our original equation of {{{log((x+21)) + log((x)) = 2}}} becomes {{{log((4+21)) + log((4)) = 2}}} which becomes {{{log((25)) + log((4)) = 2}}} which becomes {{{1.397940009 + .602059991 = 2}}} which becomes 2 = 2 so the value of x = 4 is confirmed to be good.

If x = -25, then our original equation of {{{log((x+21)) + log((x)) = 2}}} becomes {{{log((-4)) + log((-25)) = 2}}}.

since you can't take the log of a negative number (or zero) then x = -25 is not a valid answer so it is discarded.

Our answer is x = 4 which has already been confirmed to be true.

EXAMPLE NUMBER 7

Given the equation {{{log(3,(x+24)) - log(3,(x+2)) = 2}}} solve for x.

Once again, the trick is to isolate the logarithms on one side of the equation and then convert the equation to the exponential form to solve.

In this equation, rule number 2 for logarithms will be used.

That rule states that {{{log(b,(x/y)) = log(b,(x)) - log(b,(y))}}}.

Our original equation of {{{log(3,(x+24)) - log(3,(x+2)) = 2}}} becomes {{{log(3,((x+24)/(x+2))) = 2}}}

By the basic definition of logarithms {{{log(3,((x+24)/(x+2))) = 2}}} is true if and only if {{{3^2 = (x+24)/(x+2)}}} which becomes {{{9 = (x+24)/(x+2)}}}.

If we multiply both sides of the equation by (x+2) then we get {{{9*(x+2) = (x + 24)}}} which becomes {{{9x + 18 =  x + 24}}}.

If we subtract x from both sides of the equation and we subtract 18 from both sides of the equation we get {{{8*x = 6}}} which becomes {{{x = (3/4)}}} after we divide both sides of the equation by 8 and simplify.

Our answer is x = (3/4).

We substitute in the original equation of {{{log(3,(x+24)) - log(3,(x+2)) = 2}}} to get {{{log(3,((3/4)+24)) - log(3,((3/4)+2)) = 2}}} which becomes {{{log(3,(99/4)) - log(3,(11/4)) = 2}}}.

{{{log(3,(99/4)) - log(3,(11/4)) = 2}}} becomes {{{log(3,((99/4)/(11/4))) = 2}}} by rule number 2 of logarithms.
This becomes {{{log(3,((99/4)*(4/11))) = 2}}} by the rules of division which state that {{{(a/b)/(c/d)}}} is the same as {{{(a/b)*(d/c)}}}.
This becomes {{{log(3,((99)/(11))) = 2}}} after eliminating the common factors.
This becomes {{{log(3,(9)) = 2}}} after simplifying.
This is confirmed to be true because {{{log(3,(9)) = 2}}} if and only if {{{3^2 = 9}}} from the basic definition of logarithms.

Our answer is x = (3/4) which has been confirmed to be true.

EXAMPLE NUMBER 8

Since we covered rule number 1 and 2 of logarithms, we might as well round it up by covering rule number 3.

Given the equation {{{3*log(4,(x)) = 12}}} solve for x

Since the x is in the logarithmic form of the equation and not in the exponential form of the equation, this is a logarithmic equation.

Rule number 3 of logarithms states that {{{log(b,(x^n)) = n*log(b,(x))}}}

This seems to fit our example, so we take the original equation of {{{3*log(4,(x)) = 12}}} and convert it to {{{log(4,(x^3)) = 12}}}.

We use the basic definition of logarithms to transform this equation into the exponential form.

{{{log(4,(x^3)) = 12}}} is true if and only if {{{4^12 = x^3}}}

Since {{{4^12}}} = 16777216, this means that {{{x^3 = 16777216}}} which means that {{{x = root(3,16777216)}}} which means that x = 256.

If we substitute x = 256 in our original equation of {{{3*log(4,(x)) = 12}}} then we get {{{3*log(4,(256)) = 12}}} which becomes {{{log(4,(256^3)) = 12}}} which becomes {{{log(4,(16777216)) = 12}}} which is confirmed because we already know that {{{4^12 = 16777216}}}.

DERIVATION OF RULE NUMBER 1

rule number 1 states that {{{log(b,(x*y)) = log(b,(x)) + log(b,(y))}}}

let {{{b^d = x*y}}}
let {{{b^e = x}}}
let {{{b^f = y}}}

{{{x*y}}} = {{{b^d}}} = {{{(b^e)*(b^f)}}}

by the rules of exponents, {{{(b^e)*(b^f)}}} = {{{b^(e+f)}}}

this means that {{{b^d}}} = {{{b^(e+f)}}} which can only be true if {{{d = e + f}}}

by the basic definition of logarithms:
{{{b^d = (x*y)}}} if and only if {{{d = log(b,(x*y))}}}
{{{b^e = (x)}}} if and only if {{{e = log(b,(x))}}}
{{{b^f = (y)}}} if and only if {{{f = log(b,(y))}}}

The equation of {{{d = e + f}}} becomes:
{{{log(b,(x*y)) = log(b,(x)) + log(b,(y))}}} confirming the rule.

DERIVATION OF RULE NUMBER 2

rule number 2 states that {{{log(b,(x/y)) = log(b,(x)) - log(b,(y))}}}

let {{{b^d = x/y}}}
let {{{b^e = x}}}
let {{{b^f = y}}}

{{{x/y}}} = {{{b^d}}} = {{{(b^e)/(b^f)}}}

by the rules of exponents, {{{(b^e)/(b^f)}}} = {{{b^(e-f)}}}

this means that {{{b^d}}} = {{{b^(e-f)}}} which can only be true if {{{d = e - f}}}

by the basic definition of logarithms:
{{{b^d = (x/y)}}} if and only if {{{d = log(b,(x/y))}}}
{{{b^e = (x)}}} if and only if {{{e = log(b,(x))}}}
{{{b^f = (y)}}} if and only if {{{f = log(b,(y))}}}

The equation of {{{d = e - f}}} becomes:
{{{log(b,(x*y)) = log(b,(x)) - log(b,(y))}}} confirming the rule.

DERIVATION OF RULE NUMBER 3

rule number 3 states that {{{log(b,(x^n)) = n*log(b,(x))}}}

let {{{b^d = x}}}
let {{{b^e = x^n}}}

since {{{b^d = x}}} we can substitute for x in the equation of {{{b^e = x^n}}} to get:
{{{b^e = (b^d)^n}}}

by the rules of exponents, {{{(b^d)^n}}} is the same as {{{b^(d*n)}}}

this means that {{{b^e = b^(d*n)}}} which can only be true if {{{e = d*n}}}

by the basic definition of logarithms:
{{{b^d = (x)}}} if and only if {{{d = log(b,(x))}}}
{{{b^e = (x^n)}}} if and only if {{{e = log(b,(x^n))}}}

The equation of {{{e = d*n}}} becomes:
{{{log(b,(x^n)) = log(b,(x))*n}}} which is the same as:
{{{log(b,(x^n)) = n*log(b,(x))}}} confirming the rule.