When x=8, what is the value of #((x^-2)(27x^0))^(1/3)#?

Question

1969 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-24T08:30:23

#3/4#

#color(blue)(((x^-2)(27x^0))^(1/3)# #"when"# #color(brown)(x=8#

To, solve this, we need to know some exponential rules

#color(brown)(rArrx^0=1#

#color(brown)(rArrx^-z=1/(x^z)#

#color(brown)(rArrx^(1/y)=root(y)(x)#

Now, insert #8# into the problem

#rarr((8^-2)(27*8^0))^(1/3)#

Now, apply #color(brown)(x^0=1#

#rarrrarr((8^-2)(27))^(1/3)#

Apply #color(brown)(x^-z=1/x^z#

#rarr((1/8^2)(27))^(1/3)#

#rarr((1/64)(27))^(1/3)#

#rarr(27/64)^(1/3)#

Apply #color(brown)(x^(1/y)=root(y)(x)#

#rarrroot(3)(27/64)#

#rarrroot(3)((3xx3xx3)/(4xx4xx4))#

#color(green)(rArr3/4#

Hope that helps!!! ☻

Answer 2 · 2018-05-24T08:31:25

#3/4#

Given: #(x^-2*27x^0)^(1/3)#

When #x=8#, we get:

#=(8^-2*27*8^0)^(1/3)#

#=(1/8^2*27*1)^(1/3)#

#=(1/64*27)^(1/3)#

#=(27/64)^(1/3)#

#=(27^(1/3))/(64^(1/3))#

#=(root(3)27)/(root(3)64)#

#=3/4#