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Difference between Logarithms and Exponentials 
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#1
May1714, 04:23 AM

P: 62

Hello, I am beginning to learn precalculus. I understand that there are times where you can change logarithms to exponential expressions. So how are they different and similar and why are they interchangeable?



#2
May1714, 04:29 AM

PF Gold
P: 1,489

(1)##x^m=n##
(2)##m=\log_x(n)## So think it like this: (1) x is raised to the power m (Exponent) which gives n. (2) What should x be raised to get n? Answer is m. So ##y=\log_a(b)## can be changed to ##a^y=b## 


#3
May1714, 11:24 AM

P: 35

Part of the problem today in understanding the difference between logarithms and exponential is we use calculators for multiplication. Those of us from the precalculator age were introduced to logarithms at a youngish age (in my case 11). The exponential as the inverse of the logarithms came at the same time. It was 4 years of using them in maths, physics, chemistry etc. to carry out multiple multiplication and divisions that led to a thorough familiarity with them. Then I met them again in calculus particularly in the integration of x^{1}. As a result there was a natural progression to their use thereafter.



#4
May1714, 12:02 PM

P: 62

Difference between Logarithms and Exponentials
Forgive me but, x, m and n represent what exactly?



#5
May1714, 01:15 PM

PF Gold
P: 1,489

Just like in the rules of indices: ##\frac{a^m}{a^n}=a^{mn}## m and n are just numbers. 


#6
May1714, 01:27 PM

P: 62

Oh I see so it's arbitrary. so x^m=n those same arbitrary numbers can be replaced into the form m=logx^(n)? So it's interchangeable. What is the purpose of the parentheses? does that multiply out something?



#7
May1714, 01:52 PM

PF Gold
P: 1,489

It is read as "logarithms of n to base x". That's why we use' _ '. 


#8
May1814, 04:56 AM

P: 61

I would like to motivate the nature of the logarithm so that you can understand why such a device is even created. Suppose I told you that a number square is 25. Is it not fair to ask the question what is that number (whose square is 25)? And should not, much like there exists the act of squaring, an act of doing a reverse mechanism to undo the effects of squaring? It is this which in general is called an inverse operator, that which by design undoes the effects of its sibling. In that same light, the logarithm undoes the effect of the exponential Suppose you raised 5 to some number to get 25. Note that the number to be determined appears in the exponent slot and not in the base! Before it was what number square gives you 25, here it is what exponent attached to 5 gives you 25. In language we say, log base 5 of 25 = 2 (eg, the power needed to raise 5 by to achieve 25 is 2). This is called the logarithm. By design it is an inverse to exponentiating. If you were to square the square root you would have gotten back to the original value. Or even if you square root the square they would have cancelling actions. (sqrt(x))^2=sqrt (x^2)=x In here if you exponentiate the logarithm or "logarathimitiate" the exponential you will get the value you started with. x^log(x)= log(b^x)= x You would agree that these two things are equivalent: x^2=y and y= sqrt(x). One tells you that if you have knowledge on x, how to compute y while the other tells you should you have y, how to compute x. Analogously consider these two things to be equivalent: log(x)=y and x=b^y If we know x, than y is just a log away. If y is known, than a simple exponential returns us to x. The choice of which equivalent relation to use depends on the quality of our ignorance. 


#9
May1814, 08:09 AM

Mentor
P: 18,240

[tex]\sqrt{x^2} = x[/tex] Under the special condition that ##x\geq 0##, then ##x^2 = y## and ##y= \sqrt{x}## are equivalent. Not in general. 


#10
May1914, 01:33 AM

P: 61

I should have been more careful in the nature of my words or I should have resorted to some other example like the cube and cube root to avoid this. My argument was softened as it was to just motivate, at least provide a context, for understanding the logarithm. 


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