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# Justifying the logarithm properties

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.
${\mathrm{log}}_{b}\left({b}^{c}\right)=c$
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: ${\mathrm{log}}_{b}\left(MN\right)={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)$‍

Let's start by proving a specific case of the rule — the case when $M=4$, $N=8$, and $b=2$.
Substituting these values into ${\mathrm{log}}_{b}\left(MN\right)$, we see:
And so we have that ${\mathrm{log}}_{2}\left(4\cdot 8\right)={\mathrm{log}}_{2}\left(4\right)+{\mathrm{log}}_{2}\left(8\right)$.
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 $M={b}^{x}$ and $N={b}^{y}$ for some real numbers $x$ and $y$.
Then by definition, it is also true that ${\mathrm{log}}_{b}\left(M\right)=x$ and ${\mathrm{log}}_{b}\left(N\right)=y$.
Now we have:
$\begin{array}{rlrl}{\mathrm{log}}_{b}\left(MN\right)& ={\mathrm{log}}_{b}\left({b}^{x}\cdot {b}^{y}\right)& & \text{Substitution}\\ \\ & ={\mathrm{log}}_{b}\left({b}^{x+y}\right)& & \text{Properties of exponents}\\ \\ & =x+y& & {\mathrm{log}}_{b}\left({b}^{c}\right)=c\\ \\ & ={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)& & \text{Substitution}\end{array}$

## Quotient Rule: ${\mathrm{log}}_{b}\left(\frac{M}{N}\right)={\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)$‍

The proof of this property follows a method similar to the one used above.
Again, if we let $M={b}^{x}$ and $N={b}^{y}$, then it follows that ${\mathrm{log}}_{b}\left(M\right)=x$ and ${\mathrm{log}}_{b}\left(N\right)=y$.
We can now prove the quotient rule as follows:
$\begin{array}{rlrl}{\mathrm{log}}_{b}\left(\frac{M}{N}\right)& ={\mathrm{log}}_{b}\left(\frac{{b}^{x}}{{b}^{y}}\right)& & \text{Substitution}\\ \\ & ={\mathrm{log}}_{b}\left({b}^{x-y}\right)& & \text{Properties of exponents}\\ \\ & =x-y& & {\mathrm{log}}_{b}\left({b}^{c}\right)=c\\ \\ & ={\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)& & \text{Substitution}\end{array}$

## Power Rule: ${\mathrm{log}}_{b}\left({M}^{p}\right)=p{\mathrm{log}}_{b}\left(M\right)$‍

This time, only $M$ is involved in the property and so it is sufficient to let $M={b}^{x}$, which gives us that ${\mathrm{log}}_{b}\left(M\right)=x$.
The proof of the power rule is shown below.
$\begin{array}{rlrl}{\mathrm{log}}_{b}\left({M}^{p}\right)& ={\mathrm{log}}_{b}\left({\left({b}^{x}\right)}^{p}\right)& & \text{Substitution}\\ \\ & ={\mathrm{log}}_{b}\left({b}^{xp}\right)& & \text{Properties of exponents}\\ \\ & =xp& & {\mathrm{log}}_{b}\left({b}^{c}\right)=c\\ \\ & ={\mathrm{log}}_{b}\left(M\right)\cdot p& & \text{Substitution}\\ \\ & =p\cdot {\mathrm{log}}_{b}\left(M\right)& & \text{Multiplication is commutative}\end{array}$
Alternatively, we can justify this property by using the product rule.
For example, we know that ${\mathrm{log}}_{b}\left({M}^{p}\right)={\mathrm{log}}_{b}\left(M\cdot M\cdot \text{…}\cdot M\right)$, 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.
$\begin{array}{rlrl}{\mathrm{log}}_{b}\left({M}^{p}\right)& ={\mathrm{log}}_{b}\left(M\cdot M\cdot \text{…}\cdot M\right)& & \text{Definition of exponents}\\ \\ & ={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(M\right)+\text{…}+{\mathrm{log}}_{b}\left(M\right)& & \text{Product rule}\\ \\ & =p\cdot {\mathrm{log}}_{b}\left(M\right)& & \text{Repeated addition is multiplication}\end{array}$
And so you have it! We have just proven the three logarithm properties!

## Want to join the conversation?

• Why is it useful to prove these properties? Is it not sufficient to merely use them as they come up in our homework?
• 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 :-)
• how can we prove the log property which is used fr calculators?
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).
• isn’t this not more difficult than the videos?
• It depends on you
• 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.
• 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.
• Why is it useful to prove these properties? Is it not sufficient to merely use them as they come up in our homework?
• 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.
• how can we expand (log2x)^2?
• 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)
• 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?
• 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?
• It is log(base 5) and 5^ that are opposite operations, so 5^(log(base 5) x)=x and log(base 5)(5^x) = x. First is 5 to the power of log base 5 of x and second is log base 5 of 5 to the x power, it is hard to type these out without confusing parehtheses.
• Why is a^log_a(x) = x? and why does the base has to be positive?
• q1: so this statement is reached via log and exponential functions, so in the case of x not being an imaginary number:
f(x) = aˣ
the inverse is true when:
f⁻¹(f(x)) = x or f⁻¹f(x)
the inverse is logₐx (x and y swap technique), so:
a^(logₐx) = x

example:
let logₐx be b;
aᵇ = x
let a = 5 and x = 10;
5ᵇ = 10
b = log₅10
now substitute log₅10 back into 5ᵇ = 10
5^(log₅10) = 10
now return the 5 and 10 values back to a and x:
a^(logₐx) = x

q2: the base must be positive because when we rewrite a logarithmic expression with a negative base, using the base swap rule, we will be putting a negative value through log.

example:
log₋₂8= log₂8/(log₂-2)
you cannot log a negative value because it would be like trying to find a base 2 with a negative value. noo