How do you simplify #[( \frac { 1} { 2} ) ( \frac { 1} { 4} ) ^ { - ( \frac { 1} { 2} ) ^ { - 1} } + ( \frac { 1} { 25} ) ^ { - 3^ { - 1} } + ( \frac { 1} { 81} ) ^ { - 16} ] ^ { - \frac { 1} { 2} }#?

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Answer 1 · 2017-11-15T11:41:54

#sqrt(h)#

when:
#h=1/4+1/15625+81^16~~3.433*10^30~~3# trillions trillions trillions ...
(We called it "h" for "huge" : ))

Power's laws:
#(x^a)^b=x^(a*b)#
#(x^a)(x^b)=x^(a+b)#
#(x^-a)=1/(x^a)#
#(x^(1/2))=sqrt(x)#

First:
#((1/4)^(-(1/2)))^-1=((x)^a)^b#

#(x=1/4 , a=-1/2 , b=-1)#
#=> ((x)^a)^b=x^(a*b)#

#=> (1/4)^(-(1/2)*(-1))=(1/4)^(+1/2)#

#(x^(1/2))=sqrt(x)#
#=> (x=1/4)#

#=>(1/4)^(+1/2)=sqrt(1/4)=1/2#

Second:
#((1/25)^(-3))^-1=((x)^a)^b#
#(x=1/25 , a=-3 , b=-1)#
#=> ((x)^a)^b=x^(a*b)#

#=> (1/25)^(-3*(-1))=(1/25)^(+3)#
#=>(1/25)^(+3)=(1/25)*(1/25)*(1/25)=1/(25*25*25)=1/15625#

third:
#(1/81)^(-16)=(x^-a)#

#(x=1/81 , a=-16)#
#=>(x^-a)=1/(x^a)#

#=>(1/81)^(-16)=1/(1/81)^16=(1/1)/(1^16/81^16)=(1/1)/(1/81^16)=81^16/1=81^16#

forth (with help from first):
#(1/2)*((1/4)^(-(1/2)))^-1#
#=>(1/2)*(1/2)=(1*1)/(2*2)=1/4#

fifth (with help from first, second, third and forth):
#(1/2)*((1/4)^(-(1/2)))^-1+((1/25)^(-3))^-1+(1/81)^(-16)#
#=>1/4+1/15625+81^16#

Now you need to solve it, and than you got a really really really big number
lets call that number you get "h" (h for huge : ))

next step, you have:
#(h)^(-1/2)=(x^-a)#

#(x=h , a=-1/2)#
#=> (x^-a)=1/(x^a)#

#=> (h)^(-1/2)=(1/h^(1/2))#

#(x^(1/2))=sqrt(x)#
#(x=h)#

#=> 1/h^(1/2)=sqrt(h)#

So, the result of this is #sqrt(h)#,
when h= that huge number we got before