Question #98cf5

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Answer 1 · 2017-03-11T18:19:49

#k=32#

The idea here is that you can write

#4^5 = (2^2)^(5) = 2^(2 * 5) = 2^10#

and

#10^5 = (2 * 5)^5= 2^5 * 5^5#

The starting equation

#4^5 * 5^5 = k * 10^5#

now becomes

#2^10 * color(red)(cancel(color(black)(5^5))) = k * 2^5 * color(red)(cancel(color(black)(5^5)))#

which si equivalent to

#2^10 = k * 2^5#

Rearrange to solve for #k#

#k = 2^10/2^5 = 2^(10 - 5) = 2^5#

Therefore,

#k = 32#

and

#4^5 * 5^5 = 32 * 10^5#

#1024 * 3125 = 32 * 100,000#

#(3,200,000)/(100,000) = 32 " "color(darkgreen)(sqrt())#