How do you evaluate #frac { 4^ { 10} \cdot 8^ { - 3} \cdot 16^ { - 2} } { 32}#?

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Answer 1 · 2017-01-29T16:55:47

#= 1/2^2#

#=1/4#

As soon as there are indices, you cannot cancel any of the bases.

Working out each factor to its actual value is not practical - the values are far too big.

Write each base as the product of its prime factors.
You might have noticed that all the bases are powers of 2.

#color(red)(4 = 2^2)#
#color(blue)(8= 2^3)#
#color(purple)(16= 2^4)#
#color(forestgreen)(32=2^5)#

#(color(red)(4)^10*color(blue)(8)^-3*color(purple)(16)^-2)/color(forestgreen)(32)#

#=(color(red)((2^2))^10*color(blue)((2^3))^-3*color(purple)((2^4))^-2)/color(forestgreen)((2^5))#

Multiply the indices: #(x^m)^n = x^(mxxn)#

#=(color(red)(2^20)*color(blue)(2^-9)*color(purple)(2^-8))/color(forestgreen)((2^5))#

Add the indices of like bases: #x^m xx x^n = x^(m+n)#

#=2^3/2^5#

#= 1/2^2#

#=1/4#

Answer 2 · 2017-03-03T16:48:28

#1/4#

#(4^10*8^-3*16^-2)/32#

#;.=((color(red)2^2)^10*(color(red)2^3)^-3*(color(red)2^4)^-2)/color(red)2^5#

#:.=(color(red)2^20*color(red)2^-9*color(red)2^-8)/color(red)2^5#

#:.=(color(red)2^(20-9-8))/color(red)2^5#

#:.=color(red)2^3/color(red)2^5#

#:.=(cancel2^1*cancel2^1*cancel2^1)/(cancel2^1*cancel2^1*cancel2^1*2*2#

#:.=1/4#