Question #24dd4

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Answer 1 · 2018-01-30T23:48:37

Correct answer is [A]. A full explanation of how you get that result follows.

#log_c2=0.315# is essentially the same as writing #c^0.315 = 2# and #log_c3=0.500# is equivalent to #c^0.500=3# etc.

To answer this problem, find a way to create 11 250 as a product of the factors 2, 3 and 5.

#11 250 = 2 xx 3 xx 3 xx 5 xx 5 xx 5 xx 5 = 2 xx 3^2 xx 5^4#

Using the information that leads of this explanation, we can now write

#log_c11250 = log_c2+2log_c3+4log_c5#

using the fact that #log(a^x) = xloga#.

Now, substitute the values you know:

#log_c11250 = 0.315+2(0.500)+4(0.732) = 4.243#