Hard algebra/ logarithm problem

[Kungshamji]

How do you solve this?

x^0.63 = x^0.5 +1
I got it from Google, and the answer is supposed to be 9.
The equation above is just simplified. How do you solve it with algebra?

 

[Deleted member 4993]

How do you solve this?

x^0.63 = x^0.5 +1
I got it from Google, and the answer is supposed to be 9.
The equation above is just simplified. How do you solve it with algebra?

Let

log₃(2) = m

2 = 3^m

1/2 = 3^(-m)

continue.....

 

[Dr.Peterson]

How do you solve this?

x^0.63 = x^0.5 +1
I got it from Google, and the answer is supposed to be 9.
The equation above is just simplified. How do you solve it with algebra?

First, you need to be aware that in "simplifying" the equation you have changed it; an exact solution of the original will not be a solution of yours, with a rounded exponent. So you can't do that.

Second, the first trick is just to confirm that 9 is an exact solution! Khan's suggestion will help at least with that. If 3^m = 2, what is 9^m?

Third, I don't generally expect an equation like this to be solvable by algebra, though it sometimes happens if there is a suitable trick. It may be that the best way to solve it is just to graph each side, find what looks like the only solution, and check it. Algebra can't do everything, despite what algebra teachers may seem to imply!

So, the remaining question is, where did this really come from? Where "on Google" was it, and what was said about it? This is part of what we mean in asking for context.

 

[Kungshamji]

I know that the simplifying, isn’t exact. However, I just want to learn step by step how to solve the equation. Since, I know that I can plot in the values in y = x^0.5 +1
And y2 = x^log3(2),
And then just see where they intersect. I therefore want to learn how to solve it for fun. I just found the problem on google and I have tried lots of things.

 

[Deleted member 4993]

I know that the simplifying, isn’t exact. However, I just want to learn step by step how to solve the equation. Since, I know that I can plot in the values in y = x^0.5 +1
And y2 = x^log3(2),
And then just see where they intersect. I therefore want to learn how to solve it for fun. I just found the problem on google and I have tried lots of things.

Are you familiar with numerical method/s (e.g. Newton's method, bi-section method, etc.)?

 

[Kungshamji]

Are you familiar with numerical method/s (e.g. Newton's method, bi-section method, etc.)?

No, I’m not.

 

[Dr.Peterson]

You still haven't told us where the problem actually came from (what specific site) and what was said about it. That could tell us a lot about the intent of the problem.

I would say that it is probably meant to be solved "by inspection", which means by looking at the graph or otherwise finding a solution, and then convincing yourself that there is only one solution (such as by the fact that both sides of the equation are monotonically increasing, so they can't cross more than once).

But when you have a problem that you can't solve algebraically or by inspection, numerical methods are the next thing to try. Since you haven't learned them, you won't be expected to use them, but they are worth being aware of, at least. Try searching for the term!

 

[JeffM]

As has been previously explained, this type of equation cannot, in general, be solved by algebraic means. That does not mean such equations do not have solutions. If they have simple solutions, such as integer solutions, they can be found exactly by non-algebraic methods. If they have irrational solutions, a rational approximation can be found by non-algebraic means. And you can use algebra to confirm at least some solutions that algebra cannot find on its own.

[MATH]f(x) = x^{log_3(2)} \text { and } g(x) = \sqrt{x} + 1.[/MATH]
[MATH]a^{log_3(2)} = \sqrt{a} + 1 \iff f(a) = g(a).[/MATH]
[MATH]x = 9 \implies g(9) = \sqrt{9} + 1 = 3 + 1 = 4.[/MATH]
[MATH]f(x) = x^{log_3(2)} \implies log_3\{(f(x)\} = log_3(2) * log_3(x).[/MATH]
[MATH]\therefore x = 9 \implies log_3\{f(9)\} = log_3(2) * log_3(9) \implies[/MATH]
[MATH]log_3\{f(9)\} = log_3(2) * log_3(3^2) = 2 * log_3(2) * log_3(3) = log_3(2^2) * 1 = log_3(4) \implies[/MATH]
[MATH]f(9) = 4 = g(9) \implies 9^{log_3(2)} = \sqrt{9} + 1.[/MATH]
That is an algebraic proof that 9 is a solution if that is what you were after.

 

[Kungshamji]

You still haven't told us where the problem actually came from (what specific site) and what was said about it. That could tell us a lot about the intent of the problem.

I would say that it is probably meant to be solved "by inspection", which means by looking at the graph or otherwise finding a solution, and then convincing yourself that there is only one solution (such as by the fact that both sides of the equation are monotonically increasing, so they can't cross more than once).

But when you have a problem that you can't solve algebraically or by inspection, numerical methods are the next thing to try. Since you haven't learned them, you won't be expected to use them, but they are worth being aware of, at least. Try searching for the term!

A super hard logarithm problem

Hey, it's my first time visiting in a math forum, but I seriously need help. Since there are no active forums in Hebrew (I'm from Israel), I went looking for English forums, like this one. So my math teacher gave it to us today, saying he will bump up the grade of anyone solving it, showing...

mymathforum.com

^ that’s the problem. Found it looking for advanced/hard math problems, no one solves it in the forum above. I don’t know how that’s going to help me solve the equation.

 

[Kungshamji]

As has been previously explained, this type of equation cannot, in general, be solved by algebraic means. That does not mean such equations do not have solutions. If they have simple solutions, such as integer solutions, they can be found exactly by non-algebraic methods. If they have irrational solutions, a rational approximation can be found by non-algebraic means. And you can use algebra to confirm at least some solutions that algebra cannot find on its own.

[MATH]f(x) = x^{log_3(2)} \text { and } g(x) = \sqrt{x} + 1.[/MATH]
[MATH]a^{log_3(2)} = \sqrt{a} + 1 \iff f(a) = g(a).[/MATH]
[MATH]x = 9 \implies g(9) = \sqrt{9} + 1 = 3 + 1 = 4.[/MATH]
[MATH]f(x) = x^{log_3(2)} \implies log_3\{(f(x)\} = log_3(2) * log_3(x).[/MATH]
[MATH]\therefore x = 9 \implies log_3\{f(9)\} = log_3(2) * log_3(9) \implies[/MATH]
[MATH]log_3\{f(9)\} = log_3(2) * log_3(3^2) = 2 * log_3(2) * log_3(3) = log_3(2^2) * 1 = log_3(4) \implies[/MATH]
[MATH]f(9) = 4 = g(9) \implies 9^{log_3(2)} = \sqrt{9} + 1.[/MATH]
That is an algebraic proof that 9 is a solution if that is what you were after.

Not really, I want to solve it with algebra, not prove the value. How do you know you can’t solve it with algebra?

 

[JeffM]

In general, you can use algebra to find real solutions to an equation only if you can isolate x. So, for example,

[MATH]a^{f(x)} = b^{g(x)}[/MATH] is solvable in the real numbers by algebra if

[MATH]a = b,\ a \ne 1,\ a > 0[/MATH] and you can solve [MATH]f(x) - g(x)= C[/MATH] by algebra in the real numbers.

Any equation such that you cannot isolate x by algebraic means cannot be solved by algebra.

 

[pka]

Not really, I want to solve it with algebra, not prove the value. How do you know you can’t solve it with algebra?

I will not tell you that it cannot be solved by standard means.
But you must accept that standard means often includes a guess followed by a proof.
This is such a case. Look at this plot. From looking at the intersecting plots it appears that \(\displaystyle (9,4)\) is their intersection.
Lets see: \(\displaystyle \sqrt 9 +1=4\)
You must recall that: if \(\displaystyle \large{3^{\log_3(A)}=B\\\log_3(A)\log_3(3)=\log_3(B)\\\therefore A=B}\)
Now look at \(\displaystyle \large{x=9}\) so \(\displaystyle \large{9^{\log_3(2)}=(3^2)^{\log_3(2)}=(3)^{\log_3(4)}}=4\)
So we proved the guess.
If you don't like that, well as you have been told three times there is no elementary solution.
But by guess & prove there is a standard solution.

 

[Dr.Peterson]

A super hard logarithm problem

Hey, it's my first time visiting in a math forum, but I seriously need help. Since there are no active forums in Hebrew (I'm from Israel), I went looking for English forums, like this one. So my math teacher gave it to us today, saying he will bump up the grade of anyone solving it, showing...

mymathforum.com

^ that’s the problem. Found it looking for advanced/hard math problems, no one solves it in the forum above. I don’t know how that’s going to help me solve the equation.

I didn't say that telling us the source would help you solve it; but it helps us know whether you have any reason to think it can be solved, or by what means.

I see some sort-of-interesting thoughts in some of the answers there, and those that are valid seem to use "inspection" at some point. As we have said, that is a perfectly valid step.

I also see that the problem was given as a challenge, possibly either to get students to see that typically taught methods are not sufficient to solve everything, or to eventually explain that inspection is valid. Nothing there gives me any reason to believe that the sort of methods you expect will accomplish anything.