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Difference between Logarithms and Exponentials

  1. May 17, 2014 #1
    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. jcsd
  3. May 17, 2014 #2


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    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##
  4. May 17, 2014 #3
    Part of the problem today in understanding the difference between logarithms and exponential is we use calculators for multiplication. Those of us from the pre-calculator 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.
  5. May 17, 2014 #4
    Forgive me but, x, m and n represent what exactly?
  6. May 17, 2014 #5


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    It can be any number you want. I just gave a general rule.
    Just like in the rules of indices: ##\frac{a^m}{a^n}=a^{m-n}## m and n are just numbers.
  7. May 17, 2014 #6
    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?
  8. May 17, 2014 #7


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    It should be ##m=\log_x(n)## or m=log_x(n)
    It is read as "logarithms of n to base x". That's why we use' _ '.
    No. I use brackets for my own convenience. I also use brackets in trigonometric functions. It makes things neat.
  9. May 18, 2014 #8
    Hello and Welcome!

    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.
    Last edited: May 18, 2014
  10. May 18, 2014 #9


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    I disagree with this. Square root and squares are not inverses and do not always cancel eachother out. The right relation is

    [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.
  11. May 19, 2014 #10
    Yes of course, and as the square of anything induces an inherent ignorance on the sign, the square root can make no inference on that subject. In fact the square had the property of taking any real number and putting out a real number. The mapping from one to the other via the square is not injective (e.g, two distinct inputs, x and minus x, yield the same output). It was this shortcoming, I think, in the square that left no room for a genuine inverse to exist.

    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.
    Last edited: May 19, 2014
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