Hi!

 

I need some help to solve this... x = 5*log_2(x)

 

I proceed 2^x = 2^(5*log_2(x)) ==> 2^x = x^5 but I can't solve any of these cases.

 

The answer is x =1.177278550355 (approx.)

 

Thanks!

I just tried this in Mathematica and apparently it can't be solved without using this function:

 

http://mathworld.wolfram.com/LambertW-Function.html

Just through some quick numerical/graphical thinking there must be at least (and I think at most but I am uncertain) two real solutions.

 

x x^5 2^x

======================

0 0 < 1

1 1 < 2

2 32 > 4

10 10^5 > 1024

100 10^10 < 2^100

 

The two lines y=x^5 and y=2^x cross twice between 0 and 100 therefore 2^x=x^5 twice between 0 and 100. One point is clearly between 1 and 2 (and is probably your solution) the other is between 10 and 100 - I would hazard a guess around 20. I cannot see any way to solve the equation easily and suggest you try Capn suggestion

I need some help to solve this... [imath]x = 5 \cdot \log_2 (x)[/imath]

 

I proceed [imath]2^x = 2^{5 \cdot \log_2 (x)} \implies 2^x = x^5[/imath].

For one I think that should be

 

[math]\begin{array}{rcl}

x &=& 5 \cdot \log_2 (x) \\

2^x &=& 2^{5} \cdot 2^{\log_2 (x)} \\

2^x &=& 2^5 \cdot x

\end{array}[/math]

 

 

Just through some quick numerical/graphical thinking there must be at least (and I think at most but I am uncertain) two real solutions.

Yeah, at most two - [imath]y(x) = 5 \cdot \log_2 (x) - x[/imath] has just the one turning point.

I severely doubt there's a closed form expression for either.

Tree

 

x = 5.log₂(x)

x = log₂(x⁵)

2^x = 2^log₂(x⁵)

2^x = x⁵

Tree

 

x = 5.log₂(x)

x = log₂(x⁵)

2^x = 2^log₂(x⁵)

2^x = x⁵

Taking this one step further yields [math]x=2^{x/5}[/math], and this makes for a good fixed point iterator for finding that first root. The original equation, [math]x=5\log_2x[/math], makes for a good fixed point iterator for finding the second root.

 

 

 

Yep. And if you ever want to take the grind out of the calculation http://www.spearmansrank.com/fixed_point.php - works really nicely

Yep. And if you ever want to take the grind out of the calculation http://www.spearmans...fixed_point.php - works really nicely

Nice. This, however, gives me pause:

Make sure you inclose powers in brackets. For example 3x²+3x should be written as (3x^2)+3x. Without the brackets, there's no way of knowing that you didn't mean 3x^2+3x

O RLY? I suspected a problem in the parser/interpreter re precedence rules. There is no doubt as to what 3x^2+3x should calculate. Do they give exponentiation an incredibly low precedence that they feel compelled to have this caveat? Nope. They must assume that the typical user doesn't know the precedence rules.

 

 

Thank's for the answers!

Nice. This, however, gives me pause:

O RLY? I suspected a problem in the parser/interpreter re precedence rules. There is no doubt as to what 3x^2+3x should calculate. Do they give exponentiation an incredibly low precedence that they feel compelled to have this caveat? Nope. They must assume that the typical user doesn't know the precedence rules.

 

Wasn't it some newspaper baron who said that one can never over-estimate the stupidity of the general public. I reckon after the tenth email saying that their page turned out the wrong answer they realised that even people using a fixed point iterator could be incredibly dopey. On the parsing side - it couldn't work out what log2(x) meant whereas the "gold standard" wolfram alpha has no problems with it at all.

 

 

 

Wasn't it some newspaper baron who said that one can never over-estimate the stupidity of the general public. I reckon after the tenth email saying that their page turned out the wrong answer they realised that even people using a fixed point iterator could be incredibly dopey.

That's possible, particular so since their parser does appear to have precedence correct. That said, their example should have been something more reasonable and helpful such as Make sure you enclose powers in brackets. For example, the cube root of x should be written as x^(1/3) rather than x^1/3 and 3^{x^2+3x} should be written as 3^(x^2+3x).

 

On the parsing side - it couldn't work out what log2(x) meant whereas the "gold standard" wolfram alpha has no problems with it at all.

Hmm. I just assumed that it wouldn't understand log2(x) and used log(x)*5/log(2) in lieu of 5*log2(x).

 

 

 

I do not know why?

I do not know why?

 

ZSM - which bit do you not understand

it is an interest question!

 

ZSM - which bit do you not understand

 

I see! thank YOU!