Simple Proof?

[SirLazenby]

This seems like an easy question but, for the life of me, I can’t figure out how to prove x = 2 solving for (x) where...
Deductive reasoning allows one to figure it out but, how does one prove it?

 

[Steven G]

Use the change of base formula that states [math] log_ab = \dfrac {log_cb}{log_ca}[/math]

 

[pka]

This seems like an easy question but, for the life of me, I can’t figure out how to prove x = 2 solving for (x) where...
View attachment 23917Deductive reasoning allows one to figure it out but, how does one prove it?

\(x=\dfrac{\log(2x)}{\log(x)}=\dfrac{\log(2)+\log(x)}{\log(x)}=\dfrac{\log(2)}{\log(x)}+1\)
That does not help with finding the solution, but it dos make clear that \(x=2\) is the solution.

 

[Steven G]

[math]x=\dfrac{\log 2x}{log x} = \log_x2x[/math]
So x^x = 2x.

I suspect that x=2 is NOT the only solution.

 

[Harry_the_cat]

Desmos graphs y=x^x as the red curve and y=2x as blue. Seems like there are 2 solutions.
The domain is x>0 from the original equation given.

 

[Harry_the_cat]

To "prove" that x=2 is a solution you just need to show that \(\displaystyle 2=\frac {log (2*2)}{log(2)}\). You can do that by substitution.

To "solve" \(\displaystyle x=\frac{log(2x)}{log(x)}\) is a different matter.

 

[Deleted member 4993]

[math]x=\dfrac{\log 2x}{log x} = \log_x2x[/math]
So x^x = 2x.

I suspect that x=2 is NOT the only solution.

WA agrees !!

 

[SirLazenby]

View attachment 23918
Desmos graphs y=x^x as the red curve and y=2x as blue. Seems like there are 2 solutions.
The domain is x>0 from the original equation given.

Thanks for the reply Harry. Yea; I should have noted that (x) could also be 0.346 however, I still can’t figure out how to prove it on paper. I know I can use POE substitution or look for intersects on a graph but, I’d really like to know if there is a mathematical procedure I can use to prove it; it seems like there has to be. It’s kind of driving me crazy.

 

[Deleted member 4993]

Thanks for the reply Harry. Yea; I should have noted that (x) could also be 0.346 however, I still can’t figure out how to prove it on paper. I know I can use POE substitution or look for intersects on a graph but, I’d really like to know if there is a mathematical procedure I can use to prove it; it seems like there has to be. It’s kind of driving me crazy.

Why do you think - graphical procedure is not mathematical procedure.

 

[Dr.Peterson]

Thanks for the reply Harry. Yea; I should have noted that (x) could also be 0.346 however, I still can’t figure out how to prove it on paper. I know I can use POE substitution or look for intersects on a graph but, I’d really like to know if there is a mathematical procedure I can use to prove it; it seems like there has to be. It’s kind of driving me crazy.

You're confusing the words "solve" and "prove". To prove that 0.346 is (approximately) a solution, you can plug it in and see if it works. What you want is a procedure to find that solution, perhaps exactly. The answer there is, you can't! This kind of equation (transcendental -- with x both inside and outside the logs) can't be solved exactly except by something like trial and error. See post #6.

There are ways to approximate a solution, such as Newton's method, that can work for virtually any equation; but the answer will not (in general) be exact. And in fact something like that is what a graphing calculator or program will use to find the coordinates of the intersection.

We tend to teach equations for which there is a simple method, and thus fail to make students aware that algebra can't solve everything!