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## #1 2005-12-21 09:27:04

God
Member
Registered: 2005-08-25
Posts: 59

### Is there any way to solve this algebraically?

Can this be solved fully algebraically?

2^x = 2x

~ Thanks

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## #2 2005-12-21 09:38:40

mikau
Member
Registered: 2005-08-22
Posts: 1,504

### Re: Is there any way to solve this algebraically?

Wierd. The answer is obviously x =  2 but I can't seem to solve it using logarithms or anything like that.

A logarithm is just a misspelled algorithm.

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## #3 2005-12-21 09:41:55

mikau
Member
Registered: 2005-08-22
Posts: 1,504

### Re: Is there any way to solve this algebraically?

One thing I noted is if x^n = n, then x is then x is the nth root of n. (you can prove this by raising both sides to the 1/nth power. But I don't think that will help here.

A logarithm is just a misspelled algorithm.

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## #4 2005-12-21 09:48:44

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: Is there any way to solve this algebraically?

Hmm. I don't really know what I'm doing here, but I'll give it a go.

2^x = 2x
2^(x-1) = x
x-1 = lg(2) x

...and I don't know after that. Bleh.

x = 1 would also be an answer.

Why did the vector cross the road?
It wanted to be normal.

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## #5 2005-12-21 11:02:50

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

### Re: Is there any way to solve this algebraically?

What about the inverses of the functions?

I don't know what to do next.

If you combine mathsypersons stuff with mine,
you get x-1 = x/2, but I don't know if that's legal
after I took the inverse.

Last edited by John E. Franklin (2005-12-21 11:06:27)

igloo myrtilles fourmis

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## #6 2005-12-21 22:53:27

God
Member
Registered: 2005-08-25
Posts: 59

### Re: Is there any way to solve this algebraically?

Yeah I tried logs and changes of bases and etc but it just simplifies to the original equation.

The only answers are x = 1 and x = 2, which can be proved once you do "find" them...

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## #7 2005-12-21 23:09:14

ganesh
Registered: 2005-06-28
Posts: 24,639

### Re: Is there any way to solve this algebraically?

Let lg2 represent logarithm to the base 2.

2^x=2x
Taking lg2 on both sides,
x = lg2(2x) = lg2(2) + lg2(x)
lg2(2)=1,
Therefore,
x = 1+ lg2(x)
x-1 = lg2(x)
Raising both sides to the power 2,
2^(x-1)=x
For 1 and 2 alone, the LHS=RHS.
Mathsyperson was on the right path (as he's always )

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #8 2005-12-22 04:55:14

seerj
Member
Registered: 2005-12-21
Posts: 42

### Re: Is there any way to solve this algebraically?

Hi.
I tried a thing like this  (but I think that it's the same thing)
Let ln log base e
2^x=2x
ln(2^x)=ln2+lnx
xln2=ln2+lnx
x=1+ln(x)/ln(2)
and so

x=1+Log    (x)
2

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## #9 2005-12-22 07:12:08

mikau
Member
Registered: 2005-08-22
Posts: 1,504

### Re: Is there any way to solve this algebraically?

brilliant work, guys!

Logarithms rock!

A logarithm is just a misspelled algorithm.

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## #10 2005-12-22 08:13:55

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

### Re: Is there any way to solve this algebraically?

at the end of post#5 I had  x-1 = x/2, but is this allowed since
I took the inverse of the functions?

igloo myrtilles fourmis

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## #11 2005-12-28 06:37:50

God
Member
Registered: 2005-08-25
Posts: 59

### Re: Is there any way to solve this algebraically?

It is definately true that it only works for 1 and 2, but I guess, being more specific, is it possible to isolate x? As in use algebra to bring the equation down to an x = 1?

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## #12 2005-12-28 09:18:09

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: Is there any way to solve this algebraically?

x=1 and x=2.

IPBLE:  Increasing Performance By Lowering Expectations.

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## #13 2005-12-28 09:27:47

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: Is there any way to solve this algebraically?

Geometric proof:

Last edited by krassi_holmz (2005-12-28 09:53:50)

IPBLE:  Increasing Performance By Lowering Expectations.

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## #14 2005-12-28 11:25:46

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: Is there any way to solve this algebraically?

And we'll proof that there doesn't exist line that intersects with 2^i more than 2 times. That's because
ln x means natural logaritm of x
(2^i)'=2^i(ln2)=2^iC
That means that (2^i)' is a monotonic growing function so (2^i) is "convex".

IPBLE:  Increasing Performance By Lowering Expectations.

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## #15 2005-12-28 11:34:39

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: Is there any way to solve this algebraically?

Condition for existing line that divides grafhic of f(x) more than 2 times is that is nessesery to exist at least one inflex point.

IPBLE:  Increasing Performance By Lowering Expectations.

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## #16 2005-12-28 11:38:11

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: Is there any way to solve this algebraically?

Could somebody proof the upper using analisis?
I'll try solving it but i'm not so good at analisis.

IPBLE:  Increasing Performance By Lowering Expectations.

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