Study the proofs of the logarithm properties: the product rule, the quotient rule, and the power rule.

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AymeeLove

8 years agoPosted 8 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?

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(13 votes)

Bean Jaudrillard

8 years agoPosted 8 years ago. Direct link to Bean Jaudrillard's post “Sure,you could skip the p...”

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 :-)(142 votes)

Esha

8 years agoPosted 8 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?

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(15 votes)

Paul Miller

8 years agoPosted 8 years ago. Direct link to Paul Miller's post “If you are asking about t...”

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).(31 votes)

Nuha

a year agoPosted a year ago. Direct link to Nuha's post “isn’t this not more diffi...”

isn’t this not more difficult than the videos?

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(13 votes)

WillWinLiu

a year agoPosted a year ago. Direct link to WillWinLiu's post “It depends on you”

It depends on you

(7 votes)

Shubhasri Vanam

7 years agoPosted 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.•

(8 votes)

Jake D

7 years agoPosted 7 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.

(10 votes)

lauren udalor

a year agoPosted a year 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?

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(4 votes)

Oliver Gough

a year agoPosted a year 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.

(13 votes)

Sarah N.

4 years agoPosted 4 years ago. Direct link to Sarah N.'s post “how can we expand (log2x)...”

how can we expand (log2x)^2?

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(5 votes)

A/V

4 years agoPosted 4 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 simplifiedNote that:

(log(x))² ≠ 2log(x)

only that,

log(x²) = 2log(x)(12 votes)

Tyler Isaacson

3 years agoPosted 3 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?

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(7 votes)

loumast17

3 years agoPosted 3 years ago. Direct link to loumast17's post “https://www.khanacademy.o...”

https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:logs/x2ec2f6f830c9fb89:log-prop/a/properties-of-logarithms

At the top it states M,N>0, so the properties have the stipulation that the number inside a log must be positive.

(6 votes)

Simum

3 years agoPosted 3 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?

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(5 votes)

David Severin

3 years agoPosted 3 years ago. Direct link to David Severin's post “It is log(base 5) and 5^ ...”

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.

(6 votes)

Simum

3 years agoPosted 3 years ago. Direct link to Simum's post “Why is a^log_a(x) = x? a...”

Why is a^log_a(x) = x? and why does the base has to be positive?

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(5 votes)

chin

a year agoPosted a year ago. Direct link to chin's post “*q1*: so this statement i...”

**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) = xexample:

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(6 votes)

Angelino Desmet

8 years agoPosted 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.

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(3 votes)

InnocentRealist

8 years agoPosted 8 years ago. Direct link to InnocentRealist's post “It doesn't matter if a = ...”

It doesn't matter if a = 2 or not, Sayannil is correct.

But if you meant "a^m + a^n = a^(m+n)", that is correct, for any real number a, not just 2.

(9 votes)