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Algebra 2

Course: Algebra 2 > Unit 8

Lesson 3: Properties of logarithms
Math
Algebra 2
Logarithms
Properties of logarithms
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Justifying the logarithm properties

Google Classroom
Study the proofs of the logarithm properties: the product rule, the quotient rule, and the power rule.
In this lesson, we will prove three logarithm properties: the product rule, the quotient rule, and the power rule. Before we begin, let's recall a useful fact that will help us along the way.
log, start base, b, end base, left parenthesis, b, start superscript, c, end superscript, right parenthesis, equals, c
In other words, a logarithm in base b reverses the effect of a base b power!
Keep this in mind as you read through the proofs that follow.

Product Rule: log, start base, b, end base, left parenthesis, M, N, right parenthesis, equals, log, start base, b, end base, left parenthesis, M, right parenthesis, plus, log, start base, b, end base, left parenthesis, N, right parenthesis

Let's start by proving a specific case of the rule — the case when M, equals, 4, N, equals, 8, and b, equals, 2.
Substituting these values into log, start base, b, end base, left parenthesis, M, N, right parenthesis, we see:
log⁡2(4⋅8)=log⁡2(22⋅23)22=4 and 23=8=log⁡2(22+3)am⋅an=am+n=2+3log⁡b(bc)=c=log⁡2(4)+log⁡2(8)Since 2=log⁡2(4) and 3=log⁡2(8)\begin{aligned}\log_2({4\cdot 8})&=\log_2(2^2\cdot 2^3)&&\small{\gray{2^2=4\text{ and } 2^3=8}}\\ \\ &=\log_2(2^{2+3})&&\small{\gray{\text{$a^m\cdot a^n=a^{m+n}$}}}\\ \\ &=2+3&&\small{\gray{\text{$\log_b(b^c)=c$}}}\\ \\ &=\log_2(4)+\log_2(8)&&\small{\gray{\text{Since $2=\log_2(4)$ and $3=\log_2(8)$}}}\\ \end{aligned}
And so we have that log, start base, 2, end base, left parenthesis, 4, dot, 8, right parenthesis, equals, log, start base, 2, end base, left parenthesis, 4, right parenthesis, plus, log, start base, 2, end base, left parenthesis, 8, right parenthesis.
While this only verifies one case, we can follow this logic to prove the product rule in general.
Notice, that writing 4 and 8 as powers of 2 was key to the proof. So in general, we'd like M and N to be powers of the base b. To do this, we can let M, equals, b, start superscript, x, end superscript and N, equals, b, start superscript, y, end superscript for some real numbers x and y.
Then by definition, it is also true that log, start base, b, end base, left parenthesis, M, right parenthesis, equals, x and log, start base, b, end base, left parenthesis, N, right parenthesis, equals, y.
Now we have:
log⁡b(MN)=log⁡b(bx⋅by)Substitution=log⁡b(bx+y)Properties of exponents=x+ylog⁡b(bc)=c=log⁡b(M)+log⁡b(N)Substitution\begin{aligned}\log_b(MN)&=\log_b(b^x\cdot b^y)&&\small{\gray{\text{Substitution}}}\\ \\ &=\log_b(b^{x+y})&&\small{\gray{\text{Properties of exponents}}}\\ \\ &=x+y&&\small{\gray{\text{$\log_b(b^c)=c$}}} \\\\ &=\log_b(M)+\log_b(N)&&\small{\gray{\text{Substitution}}} \end{aligned}

Quotient Rule: log, start base, b, end base, left parenthesis, start fraction, M, divided by, N, end fraction, right parenthesis, equals, log, start base, b, end base, left parenthesis, M, right parenthesis, minus, log, start base, b, end base, left parenthesis, N, right parenthesis

The proof of this property follows a method similar to the one used above.
Again, if we let M, equals, b, start superscript, x, end superscript and N, equals, b, start superscript, y, end superscript, then it follows that log, start base, b, end base, left parenthesis, M, right parenthesis, equals, x and log, start base, b, end base, left parenthesis, N, right parenthesis, equals, y.
We can now prove the quotient rule as follows:
log⁡b(MN)=log⁡b(bxby)Substitution=log⁡b(bx−y)Properties of exponents=x−ylog⁡b(bc)=c=log⁡b(M)−log⁡b(N)Substitution\begin{aligned}\log_b\left(\dfrac{M}{N}\right)&=\log_b\left(\dfrac{b^x}{ b^y}\right)&&\small{\gray{\text{Substitution}}}\\ \\ &=\log_b(b^{x-y})&&\small{\gray{\text{Properties of exponents}}}\\ \\ &=x-y&&\small{\gray{\text{$\log_b(b^c)=c$}}}\\ \\ &=\log_b(M)-\log_b(N)&&\small{\gray{\text{Substitution}}} \end{aligned}

Power Rule: log, start base, b, end base, left parenthesis, M, start superscript, p, end superscript, right parenthesis, equals, p, log, start base, b, end base, left parenthesis, M, right parenthesis

This time, only M is involved in the property and so it is sufficient to let M, equals, b, start superscript, x, end superscript, which gives us that log, start base, b, end base, left parenthesis, M, right parenthesis, equals, x.
The proof of the power rule is shown below.
log⁡b(Mp)=log⁡b((bx)p)Substitution=log⁡b(bxp)Properties of exponents=xplog⁡b(bc)=c=log⁡b(M)⋅pSubstitution=p⋅log⁡b(M)Multiplication is commutative\begin{aligned}\log_b\left(M^p\right)&=\log_b(\left({b^x}\right)^p)&&\small{\gray{\text{Substitution}}}\\ \\ &=\log_b(b^{xp})&&\small{\gray{\text{Properties of exponents}}}\\ \\ &=xp&&\small{\gray{\text{$\log_b(b^c)=c$}}}\\ \\ &=\log_b(M)\cdot p&&\small{\gray{\text{Substitution}}}\\ \\ &=p\cdot \log_b(M)&&\small{\gray{\text{Multiplication is commutative}}} \end{aligned}
Alternatively, we can justify this property by using the product rule.
For example, we know that log, start base, b, end base, left parenthesis, M, start superscript, p, end superscript, right parenthesis, equals, log, start base, b, end base, left parenthesis, M, dot, M, dot, point, point, point, dot, M, right parenthesis, where M is multiplied by itself p times.
We can now use the product rule along with the definition of multiplication as repeated addition to complete the proof. This is shown below.
log⁡b(Mp)=log⁡b(M⋅M⋅...⋅M)Definition of exponents=log⁡b(M)+log⁡b(M)+...+log⁡b(M)Product rule=p⋅log⁡b(M)Repeated addition is multiplication\begin{aligned} \log_b(M^p) &=\log_b(M\cdot M\cdot ...\cdot M)&&\small{\gray{\text{Definition of exponents}}}\\ \\ &= \log_b(M)+\log_b(M)+...+\log_b(M)&& \small{\gray{\text{Product rule}}}\\\\ &= p\cdot \log_b(M) &&\small{\gray{\text{Repeated addition is multiplication}}}\end{aligned}
And so you have it! We have just proven the three logarithm properties!

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  • leafers ultimate style avatar for user AymeeLove
    Posted 7 years ago. Direct link to AymeeLove's post “Why is it useful to prove...”
    Why is it useful to prove these properties? Is it not sufficient to merely use them as they come up in our homework?
    (11 votes)
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    • leaf red style avatar for user Bean Jaudrillard
      Posted 7 years ago. Direct link to Bean Jaudrillard's post “Sure,you could skip the p...”
      Incredible Answer
      Great Answer
      Good Answer
      Sure,you could skip the proofs and just use these properties as they come up in your homework,but would you not like to know that how do these properties work?
      Human beings from the very beginning have been curious about how and why stuff works,and I believe that it is only because of their curiosity to know more about how and why their universe works,that they have been able to develop and evolve unlike other animals.
      Moreover, mathematics is a discipline that,along with the sciences, focuses on the how and why questions,which is the reason why we proof things,so that everyone comes to know that whatever that is in mathematics is how the universe works,and nothing is someone's made up rule,that you need to mindlessly follow.
      Anyway it's always good to know these kind of proofs because you may be asked to proof them,or something else closely related to them in certain maths tests that you may have to take :-)
      (101 votes)
  • leafers ultimate style avatar for user Esha
    Posted 7 years ago. Direct link to Esha's post “how can we prove the log ...”
    how can we prove the log property which is used fr calculators?
    (15 votes)
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    • starky ultimate style avatar for user Paul Miller
      Posted 7 years ago. Direct link to Paul Miller's post “If you are asking about t...”
      Great Answer
      Good Answer
      If you are asking about the base change formula
      log base a of b = log base c of b / log base c of a
      Which I will prove using log base e for ease of notation (Log base e of x = ln (x)) the main thing is to start where the other proofs started...the definition of the logarithm and the properties of exponents.
      Suppose I have two positive real numbers a and b, a<>1 and
      log base a of b = M.
      then I can write
      b = a^M by the definition of the logarithm.
      Now take the natural logarithm (or other base if you want) of both sides of the equation to get the equivalent equation
      ln(b)=ln(a^M).
      Now we can use the exponent property of logarithms we proved above to write
      ln(b)=M*ln(a).
      Divide both sides by ln(a) to get
      ln(b)/ln(a) = M
      Substituing in for M in our original equation we now see that
      log base a of b = M = ln(b)/ln(a).
      (26 votes)
  • aqualine sapling style avatar for user Shubhasri Vanam
    Posted 7 years ago. Direct link to Shubhasri Vanam's post “How can you prove the pow...”
    How can you prove the power rule using the power rule?
    logb(b^c)=c is actually what you used and this is actually the power rule.
    (9 votes)
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    • aqualine tree style avatar for user Jake D
      Posted 6 years ago. Direct link to Jake D's post “log_b(b^c) = c is *not* t...”
      log_b(b^c) = c is not the power rule. The power rule stated with those variables is:

      log_b(b^c) = c * log_b(b)

      But what is log_b(b)? That would be the same as asking "What exponent can we raise b to obtain b?". The answer, is obviously 1 because b^1 = b.

      The power rule more generally, can be expressed as:

      log_a(b^c) = c * log_a(b)

      So back to the original. How can we show that log_b(b^c) = c without using the power rule?

      Let's start by rewriting this as an exponent:

      log_b(b^c) = x ---> b^x = b^c

      Because b^x = b^c, x equals c. What is x? x is log_b(b^c).

      Therefore, log_b(b^c) = c. Here, we didn't use the power rule.
      (9 votes)
  • blobby green style avatar for user Nuha
    Posted 4 months ago. Direct link to Nuha's post “isn’t this not more diffi...”
    isn’t this not more difficult than the videos?
    (8 votes)
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  • blobby green style avatar for user Sarah N.
    Posted 3 years ago. Direct link to Sarah N.'s post “how can we expand (log2x)...”
    how can we expand (log2x)^2?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • leaf orange style avatar for user A/V
      Posted 3 years ago. Direct link to A/V's post “I am unsure of which vari...”
      I am unsure of which variation you are showing, so I'll show both:

      Scenario 1.) (log2(x))²
      When you have a log that is squared outside of the parenthesis, it would look like this: log²2(x)

      Using log(b)/log(a) rule to put everything as the same base, we now have:
      >> log²(x)/log²(2), which is fully simplified

      Scenario 2.) (log(2x))²
      Rewritten as like scenario 1,
      >> log²(2x) which is already simplified

      Note that:
      (log(x))² ≠ 2log(x)
      only that,
      log(x²) = 2log(x)
      (9 votes)
  • duskpin ultimate style avatar for user Angelino Desmet
    Posted 8 years ago. Direct link to Angelino Desmet's post “a^m * b^n = a^(m+n) Is i...”
    a^m * b^n = a^(m+n)

    Is it allowed to state the equation above, without explicitly mentioning a=2? Because the equation only holds when a=b.
    (3 votes)
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  • blobby green style avatar for user lauren udalor
    Posted 8 months ago. Direct link to lauren udalor's post “Why is it useful to prove...”
    Why is it useful to prove these properties? Is it not sufficient to merely use them as they come up in our homework?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • leafers sapling style avatar for user Oliver Gough
      Posted 4 months ago. Direct link to Oliver Gough's post “In math, especially advan...”
      In math, especially advanced math, the most important thing is proving things. That is how new math properties are known to be true. In school you learn geometry and one reason is that it is a nice way to learn how proofs work. Any time you are told a new math rule, you should realise that there is a proof behind it to justify why it is a rule - and it often helps you understand a rule to see it proved or even to prove it yourself.
      (9 votes)
  • aqualine ultimate style avatar for user Simum
    Posted 2 years ago. Direct link to Simum's post “How can we just write bot...”
    How can we just write both sides as exponents? How is the value of both sides equal when we take the log at the same base of sides? Is there an inverse log that cancels them?
    (3 votes)
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  • blobby green style avatar for user chloe.carman
    Posted 6 years ago. Direct link to chloe.carman's post “What is e^4 and how to yo...”
    What is e^4 and how to you find it?
    (3 votes)
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  • blobby green style avatar for user Tyler Isaacson
    Posted 2 years ago. Direct link to Tyler Isaacson's post “log(-5)-log(-1) = undefin...”
    log(-5)-log(-1) = undefined (10 to the power of anything will never be negative)

    log(-5/-1) = 0.0969100130081

    log(-5/-1) =/= log(-5)-log(-1)


    log(-x/-b) =/= log(-x)-log(-b)
    log(x/b) =/= log(x)-log(b) ?

    How does the quotient property work if a double negative cancels in one form but is undefined in the other form?
    (4 votes)
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