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

Can this be solved fully algebraically?

2^x = 2x

~ Thanks

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

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

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

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

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

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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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,334

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

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

brilliant work, guys!

Logarithms rock!

A logarithm is just a misspelled algorithm.

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

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

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

We have two answers:

x=1 and x=2.

IPBLE: Increasing Performance By Lowering Expectations.

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

Geometric proof:

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

IPBLE: Increasing Performance By Lowering Expectations.

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

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

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

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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