# B How to solve this specific logarithm problem

1. Feb 14, 2017

### srfriggen

(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. Feb 14, 2017

### Staff: Mentor

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 $5^x = 4^x + 1 \Longrightarrow x=1$.

3. Feb 14, 2017

### srfriggen

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. Feb 14, 2017

### Staff: Mentor

fresh_42 is, I believe, steering you toward using methods of calculus to show that the function has only one zero.

5. Feb 14, 2017

### srfriggen

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. Feb 14, 2017

### Staff: Mentor

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. Feb 14, 2017

### srfriggen

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. Feb 14, 2017

### Staff: Mentor

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.

9. Feb 14, 2017

### Staff: Mentor

10. Feb 15, 2017

### John Park

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. Feb 15, 2017

### jbriggs444

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
12. Feb 18, 2017

### John Park

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. Feb 19, 2017

### Staff: Mentor

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

14. Feb 22, 2017

### John Park

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.