How to solve this specific logarithm problem

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  • #1
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(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?
 

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  • #2
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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##.
 
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  • #3
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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##.
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.
 
  • #4
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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##.
fresh_42 is, I believe, steering you toward using methods of calculus to show that the function has only one zero.
 
  • #5
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fresh_42 is, I believe, steering you toward using methods of calculus to show that the function has only one zero.
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?
 
  • #6
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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?
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##
 
  • #7
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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##
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.
 
  • #8
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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
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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.
 
  • #11
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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.
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.
 
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  • #12
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That part of the argument is not well stated.
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.
 
  • #13
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Don't know if this will help, but . . .

You have x = log5( 4x+ 1) . This implies x > 0.
The equation above does not imply that x > 0. A log function can evaluate to a negative number; for example, ##\log(1/10) = -1##.
John Park said:
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.
 
  • #14
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Don't know if this will help, but . . . You have x = log5( 4x+ 1) . This implies x > 0.

The equation above does not imply that x > 0. A log function can evaluate to a negative number; for example, log ( 1 / 10 ) = − 1 log 1 10 1 .
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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