Question #98cf5

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Answer 1 · 2017-03-11T18:19:49

$k = 32$ $k=32$

The idea here is that you can write

$4^{5} = {(2^{2})}^{5} = 2^{2 \cdot 5} = 2^{10}$ $4^5 = (2^2)^(5) = 2^(2 * 5) = 2^10$

and

$10^{5} = {(2 \cdot 5)}^{5} = 2^{5} \cdot 5^{5}$ $10^5 = (2 * 5)^5= 2^5 * 5^5$

The starting equation

$4^{5} \cdot 5^{5} = k \cdot 10^{5}$ $4^5 * 5^5 = k * 10^5$

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

$4^5 * 5^5 = 32 * 10^5$

$1024 * 3125 = 32 * 100,000$

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