YAHT: Math Question

[AgaBoogaBoo]

log z = sqrrt(log z)

sqrrt = square root of

I understand that the right side can be written as log z ^ (1/2), but I don't know what to do from there. Bases are equal at that point, so their powers should be as well, but log z doesn't have an indicated power.

 

[Crusty]

log z = sqrt(log z )
log z = 1/2logz

Maybe that will help

 

[Sukhoi]

Aren't z=0, z=1 the answers? You get log z = (log z)^(1/2). Then raise each side as a power of 10 to get z = z^(1/2). Aren't the answers to that 0,1?

 

[AgaBoogaBoo]

The answer is 1, 10

 

[RaynorWolfcastle]

substitute w = sqrt(log(z)), then solve for w. Then solve for z using your 2 solutions for w (since your eqn in terms of w qill be a quadratic)

 

[eLiu]

Originally posted by: Sukhoi
Aren't z=0, z=1 the answers? You get log z = (log z)^(1/2). Then raise each side as a power of 10 to get z = z^(1/2). Aren't the answers to that 0,1?

Uhm...almost. Since you upped the powers, you need to do the same to your solution: 10^0 and 10^1 (1 and 10) are the answers here.

 

[Crypticburn]

Originally posted by: Sukhoi
Aren't z=0, z=1 the answers? You get log z = (log z)^(1/2). Then raise each side as a power of 10 to get z = z^(1/2). Aren't the answers to that 0,1?

When you raise each side to the logarithmic base you get:

10^log(z) = 10^sqrt(log(z))
z = 10^sqrt(log(z))

The right hand side of the equation does not simplify to what you suggest.

sqrt(log(z)) != log(sqrt(z)) ==> 10^sqrt(log(z)) != sqrt(z)
1/2 * log(z) = log(sqrt(z)) ==> 10^(1/2 * log(z)) == sqrt(z)

RaynorWolfcastle suggested the correct method, substitution.

Crypticburn

 

[Muzzan]

Originally posted by: Sukhoi
Aren't z=0, z=1 the answers? You get log z = (log z)^(1/2). Then raise each side as a power of 10 to get z = z^(1/2). Aren't the answers to that 0,1?

log(0) isn't defined, so z = 0 is not a solution.

 

[Sukhoi]

Oh boo, rain on my parade.