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B How to solve this specific logarithm problem

  1. Feb 14, 2017 #1
    (This is absolutely not a HW problem) I had posted in a much older thread, so just wanted to post again in a new one. How would you solve, for x:

    5x = 4x + 1

    x = 1 is clearly an answer, but I get struck when trying to solve using logs.

    If I take log5 of both sides I get,

    x = log5 ( 4x+ 1) and of course you don't distribute the log function. So how does one solve this?
     
  2. jcsd
  3. Feb 14, 2017 #2

    fresh_42

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    Why did you re-post it?

    You know that ##x=1## is a zero of ##f(x)=5^x-4^x-1##. What methods do you know to show, that it is the only zero?
    So in doing so, you know indirectly ##5^x = 4^x + 1 \Longrightarrow x=1##.
     
  4. Feb 14, 2017 #3
    Sorry to repost. I haven't posted in PF in a while and didn't realize the old post got moved to the top.

    I've been trying, and in that form too, and I just don't see it. I don't think it's in quadratic form so I don't see a way to do a u-substitution, and I can't see how I can apply the zero product law without factoring. I'm thinking I must be rusty on some exponent law that I'm missing.
     
  5. Feb 14, 2017 #4

    Mark44

    Staff: Mentor

    fresh_42 is, I believe, steering you toward using methods of calculus to show that the function has only one zero.
     
  6. Feb 14, 2017 #5
    I am not at that level. I have not done calculus in such a long time. The only thing I can think of is the mean value theorem will tell you something about whether a zero exists, but I simply do not know the method to calculate the zero. Can someone please show me how this would be solved?
     
  7. Feb 14, 2017 #6

    fresh_42

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    It cannot be solved explicitly, will say, there is no formula to get from ##5^x=4^x+1## to ##x=1##.
    Can you differentiate it? This would help to show whether the function ##f(x)=5^x-4^x-1## is increasing. Or you might be able to show, that ##f(x)<0## for ##x<1## and ##f(x)>0## for ##x>1##. These are possible ways to show that only ##x=1## solves the equation. All of them are kind of indirect, because you cannot rearrange the equation to a form ##x= \ldots##
     
  8. Feb 14, 2017 #7
    Would this be a situation for the Squeeze Theorem? It's definitely always increasing and I recall the mean value theorem would prove that there exists a real zero, but I don't know how to get 1 as an answer. Like I mentioned, I haven't done calculus in 6 years and am just grasping at straws. If it can't be done with logs I guess I just feel better I wasn't missing anything.
     
  9. Feb 14, 2017 #8

    fresh_42

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    You get ##1## as an answer only by numeric methods, e.g. the Newton-Paphson method or by using Wolframalpha or in this case by an easy guess. I think in this case it is easier to show where ##f(x) < 0## and where ##f(x) > 0## since we know the answer already than to use any theorems.
     
  10. Feb 14, 2017 #9

    jedishrfu

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  11. Feb 15, 2017 #10
    Don't know if this will help, but . . .

    You have x = log5( 4x+ 1) . This implies x > 0.

    In this range, the slope of 5^x is always greater than that of 4^x. (This can be proved by calculus, but it's probably obvious.)

    Both are monotonically increasing functions. So, geometrically, it follows that the curve of y = 5^x can't intersect that of y = 4^x + 1 more than once.
     
  12. Feb 15, 2017 #11

    jbriggs444

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    That part of the argument is not well stated.

    Both f(x) = 10x + sin x and g(x) = 10 x are monotone increasing. Yet the graphs of f(x) and g(x) intersect infinitely many times.

    A better observation is that the slope f(x) = 5x is [for positive x] greater than that of g(x) = 4x + 1 (and that are both defined for positive x). This means that the slope of f(x) - g(x) in this range is always positive and consequently that f(x) - g(x) is monotone increasing for positive x. This means that it can have at most one positive zero.
     
    Last edited: Feb 15, 2017
  13. Feb 18, 2017 #12
    I did make the point about the slopes; probably I should have repunctuated to emphasise it.

    But perhaps the main point is that in this case, as for many real-world problems, there is no clean "algebraic" way to get to the solution. You either happen to see a solution (as here) and convince yourself it's the one want, or you use some numerical approach or possibly algebraic approximation.
     
  14. Feb 19, 2017 #13

    Mark44

    Staff: Mentor

    The equation above does not imply that x > 0. A log function can evaluate to a negative number; for example, ##\log(1/10) = -1##.
     
  15. Feb 22, 2017 #14
    Look at the original equation:
    x=log5(4x+1) (Sorry for the typo above.)

    The quantity 4x must be greater than zero; therefore the argument of the log is greater than 1, and the log is positive.
     
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