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Prove logb(x) = y is equivalent to b^y = x.

If logb(x) = 5, then logb(x^2) is equal to?

TBH, I'm more interested to see how people prove the log law (second Q. is just bringing down exponent).
 
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Sy123

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They are just 'proofs'
Bringing the exponent down is sufficient
 

D94

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"Prove" is an interesting question, because log laws need to hold true to their definitions.

"Definition: Suppose that b is a positive real number not equal to one, y is a rational number and x = by. Then logbx is defined by the formula: y = logbx ." (UNSW Calculus) (but I've changed the variables to suit the question)

You could use the change of base rule to determine their equivalence, but then we use the law which we are proving to determine the answer.

The problem is that we must assume the laws hold true in maths in order to apply them. Since these are facts, they don't require "proofs", so the question is asking for some tautological reasoning, as how I have interpreted it.
 

Carrotsticks

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"Prove" is an interesting question, because log laws need to hold true to their definitions.

"Definition: Suppose that b is a positive real number not equal to one, y is a rational number and x = by. Then logbx is defined by the formula: y = logbx ." (UNSW Calculus) (but I've changed the variables to suit the question)

You could use the change of base rule to determine their equivalence, but then we use the law which we are proving to determine the answer.

The problem is that we must assume the laws hold true in maths in order to apply them. Since these are facts, they don't require "proofs", so the question is asking for some tautological reasoning, as how I have interpreted it.
Not really.

For example most people take this for granted as a definition:



But that still has to be proven (many ways of proving, one of which is the Cauchy Product), as trivial as it may be.
 

qwerty44

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Not really.

For example most people take this for granted as a definition:



But that still has to be proven (many ways of proving, one of which is the Cauchy Product), as trivial as it may be.
But isn't that different to proving logb(x)=y then b^y=x?

Logs were invented by John Napier as assistance in multiplication, and were essentially simple calculators. How can you prove a concept created by man, not discovered by man?

(Unless I am dead wrong, if so im sorry)
 

D94

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Not really.

For example most people take this for granted as a definition:



But that still has to be proven (many ways of proving, one of which is the Cauchy Product), as trivial as it may be.
But that's not a definition. We don't "define" , but we "define" the log question above as what I said previously (unless you now say the logic in my quote is wrong). Also, what I said was in the context of log laws.

When I say "don't require proofs", I mean, you are to assume it holds true, otherwise, we get into a tautological reasoning. (and yes, we then go into the truth argument which everyone would rather avoid)
 

Fus Ro Dah

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But that's not a definition. We don't "define" , but we "define" the log question above as what I said previously (unless you now say the logic in my quote is wrong). Also, what I said was in the context of log laws.

When I say "don't require proofs", I mean, you are to assume it holds true, otherwise, we get into a tautological reasoning. (and yes, we then go into the truth argument which everyone would rather avoid)
No, I'm afraid you are mistaken. We define the Logarithmic function to be the inverse of its equivalent exponential function, but there is no guarantee that it immediately implies an exponential form immediately.

For simplicity, I will use the functions and its equivalent inverse, though of course I can use any positive number 'a' in the expression . I'll only be dealing with positive 'a' because I'm assuming we are just working with the reals here.

Define a function , which we already know is very well defined. Since this is a bijective function, where function composition of and causes negation. We will call its inverse g such that and conversely

Now consider the expression which is equivalent to because we defined the logarithmic function to be the inverse g, nothing more. Making both sides the exponent of e gives us but as we have shown above, the function composition of f and g or visa versa causes negation, and so we acquire , which is what we wanted to have.
 

D94

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To be fair, I think you should reread my quoted definition. That's what this whole thread is about.

One question, where is it defined that the logarithm function is the "inverse" of the equivalent exponential function? Aren't you really paraphrasing my quote? (did you even read that?)
 

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