Help with Logarithms: Find x When 2lnx=xln2

[James889]

I have 2lnx = xln2
where x\ne2

if you start by dividing both sides by ln2
is the following legal?

\frac{2lnx}{ln2} \rightarrow x = 2ln(x-2)

e^{2ln(x-2)} = (x-2)^2

x = (x-2)^2 \implies x = 4

 

[Mentallic]

I'm confused at what you've done.

Did you manipulate \frac{2lnx}{ln2} and turn it into 2ln(x-2)? Because this is generally not correct.

From what I can gather, you've just fluked your way into breaking a rule but still finding a solution for x. That equation isn't simple to solve.

 

[James889]

Hm,

\frac{ln~x}{ln~y} \implies ln(x-y)

?

 

[yungman]

I have 2lnx = xln2
where x\ne2

if you start by dividing both sides by ln2
is the following legal?

\frac{2lnx}{ln2} \rightarrow x = 2ln(x-2)

e^{2ln(x-2)} = (x-2)^2

x = (x-2)^2 \implies x = 4

2lnx = xln2\Rightarrow \frac{d[2lnx]}{dx}=ln2\frac{dx}{dx}\Rightarrow \frac{2}{x}=ln2\Rightarrow x=\frac{2}{ln2}

 

[James889]

You miss the left hand side.e^{x}=(x-2)^2

2lnx = xln2\Rightarrow \frac{d[2lnx]}{dx}=ln2\frac{dx}{dx}\Rightarrow \frac{2}{x}=ln2\Rightarrow x=\frac{2}{ln2}

Hm, no!

 

[yungman]

Hm, no!

Then tell me what did I do wrong. You follow step by step what I did?

 

[Bohrok]

Hm,

\frac{ln~x}{ln~y} \implies ln(x-y)

?

No, you have it completely backwards...
\ln\left(\frac{x}{y}\right) = \ln x - \ln y

but this isn't going to help actually solve the equation.

 

[Redbelly98]

Hint: review the rules for manipulating logs. One of the says that

A ln(B) = ____?​

 

[yungman]

No, you have it completely backwards...
\ln\left(\frac{x}{y}\right) = \ln x - \ln y

but this isn't going to help actually solve the equation.

I got fool too! And I destroy all the evidence already!


I did it differently and I still don't see why he claimed my answer in #4 was wrong!

 

[Mentallic]

I got fool too! And I destroy all the evidence already!


I did it differently and I still don't see why he claimed my answer in #4 was wrong!

He claimed it was wrong because it is wrong...

You took the derivative and solved for that. Yeah you found the turning point, but not a root of the original equation.

 

[ideasrule]

You seem to be thinking that d[2lnx]/dx = 2lnx. It's not. It's equal to 2/x.

 

[Mentallic]

No no, yungman never asserted that. What he did was correct, but only for finding the turning point of y=2lnx-xln2.

It's kind of confusing with all the arrows and equals signs, so I'll clear it up:

2lnx = xln2

\frac{d[2lnx]}{dx}=ln2\frac{dx}{dx}

\frac{2}{x}=ln2

x=\frac{2}{ln2}

But again, it doesn't help finding the solution x=4 to this problem (which btw, the OP found by an illegal move)

 

[cap.r]

if the problem is just as simple as solving the equation for x then just raise everything to the e.

e^(lnx)^2=e^(ln2)^x this trivializes and is simple to solve.

 

[Mentallic]

if the problem is just as simple as solving the equation for x then just raise everything to the e.

e^(lnx)^2=e^(ln2)^x this trivializes and is simple to solve.

Yes so you have x^2=2^x but really, this isn't as simple to solve as you claim. It's just that we are lucky enough to have nice integer solutions for x>0. If we ignored the domain on the original equation and tried to instead solve this one, you can only get a numerical approximation for the solution in -1<x<0

 

[yungman]

No no, yungman never asserted that. What he did was correct, but only for finding the turning point of y=2lnx-xln2.

It's kind of confusing with all the arrows and equals signs, so I'll clear it up:

2lnx = xln2

\frac{d[2lnx]}{dx}=ln2\frac{dx}{dx}

\frac{2}{x}=ln2

x=\frac{2}{ln2}

But again, it doesn't help finding the solution x=4 to this problem (which btw, the OP found by an illegal move)

You loss me about "turning point"! What is a turning point?

Who said that the answer is x=4? James concluded x=4 with a wrong assumption that \frac{lnx}{ln2}=ln(x-2).

I conclude x=\frac{2}{ln2}...which is a constant number.

 

[yungman]

Yes so you have x^2=2^x but really, this isn't as simple to solve as you claim. It's just that we are lucky enough to have nice integer solutions for x>0. If we ignored the domain on the original equation and tried to instead solve this one, you can only get a numerical approximation for the solution in -1<x<0

Yes I tried that and didn't go no where!

 

[Mentallic]

You loss me about "turning point"! What is a turning point?

Who said that the answer is x=4? James concluded x=4 with a wrong assumption that \frac{lnx}{ln2}=ln(x-2).

I conclude x=\frac{2}{ln2}...which is a constant number.

You know about derivatives but not turning points?

But the answer is x=4. Just try it (and btw, 4 is a constant number too )

xln2=2lnx

4ln2=2ln4

RHS=2ln(2^2)=4ln2=LHS

So x=4 is a solution. And x=2 is too, but not 2/ln2.

 

[yungman]

You know about derivatives but not turning points?

But the answer is x=4. Just try it (and btw, 4 is a constant number too )

xln2=2lnx

4ln2=2ln4

RHS=2ln(2^2)=4ln2=LHS

So x=4 is a solution. And x=2 is too, but not 2/ln2.

I self study most of my calculus! Things sounds common to you might not be to me! The only other class I took was ODE and I have people laughing at me on some stuff too! But I was the first in class for the semester.

Do you mean the point that the function turn from a convex to a concave? I remember I read something about it a few years ago. But still never heard of turning point!

I still don't get why my method don't produce the answer. I know I got that by taking the derivative on both side, I have seen work problems using this method. Do I have to keep bitting my nails and wait until James come up with the answer??

 

[vela]

I still don't get why my method don't produce the answer. I know I got that by taking the derivative on both side, I have seen work problems using this method. Do I have to keep bitting my nails and wait until James come up with the answer??

It's because f(x)=0 and f'(x)=0 generally have different solutions. When you differentiated, you got a new equation, and its solution is not a solution to the original equation.

 

[yungman]

It's because f(x)=0 and f'(x)=0 generally have different solutions. When you differentiated, you got a new equation, and its solution is not a solution to the original equation.

Thanks, I got it.



This is where I am at:

xln2=2lnx\Rightarrow x=\frac{2}{ln2}x

\Rightarrow \frac{x}{lnx}=\frac{2}{ln2}

It is easy to see x=2. I still don't see x=4.

 

[HallsofIvy]

The original equation was 2ln(x)= x ln(2). I don't see where anyone has yet pointed out that this equation cannot be solved in terms of elementary functions. It can, of course, be written as ln(x^2)= ln(2^x) and then x^2= 2^x and that looks like, with some manipulation, it could be solved in terms of "Lambert's W function".

 

[Mentallic]

I don't see where anyone has yet pointed out that this equation cannot be solved in terms of elementary functions. It can, of course, be written as ln(x^2)= ln(2^x) and then x^2= 2^x and that looks like, with some manipulation, it could be solved in terms of "Lambert's W function".

These were my feable attempts to do so:

...That equation isn't simple to solve.

...you can only get a numerical approximation for the solution in -1<x<0 (for the equation x^2=2^x)

Yungman, a turning point is the point at which the gradient of the curve becomes zero. i.e. if the curve has a negative gradient but at some point it "flattens out" with 0 gradient and either becomes positive or negative again after that.

e.g. the curve y=x(x-1) has roots at x=0,1 but it has a turning point at x=0.5 which you can find by taking the derivative and solving for when f'(x)=0.
This is what you did previously.

 

[Redbelly98]

EDIT: Never mind! Clearly x>0 is required in the equation given in the OP.

[STRIKE]I'm not quite sure why there is all this discussion of f'(x)=0. We have an equation to solve for x, as given in the OP. Discounting the x=2 solution, there are two more values of x that solve the equation. We have found x=4, but there is one more solution to find.[/STRIKE]

[STRIKE]x²=2^x[/STRIKE]​

[STRIKE]It looks like a numerical approach is needed to find the negative solution. James889, what numerical techniques (if any) have your professor discussed?[/STRIKE]