Question

Question #24dd4

Answer 1

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

Explanation:

`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`