<span>E = {x|x is a perfect square between 1 and 9} = {1,4
</span><span>F = {x|x is an even number greater than or equal to 2 and less than 9}
</span><span>D = {x|x is a whole number}
</span>
answer
D ∩ (E ∩ F) = 4
Use the rules of logarithms and the rules of exponents.
... ln(ab) = ln(a) + ln(b)
... e^ln(a) = a
... (a^b)·(a^c) = a^(b+c)
_____
1) Use the second rule and take the antilog.
... e^ln(x) = x = e^(5.6 + ln(7.5))
... x = (e^5.6)·(e^ln(7.5)) . . . . . . use the rule of exponents
... x = 7.5·e^5.6 . . . . . . . . . . . . use the second rule of logarithms
... x ≈ 2028.2 . . . . . . . . . . . . . use your calculator (could do this after the 1st step)
2) Similar to the previous problem, except base-10 logs are involved.
... x = 10^(5.6 -log(7.5)) . . . . . take the antilog. Could evaluate now.
... = (1/7.5)·10^5.6 . . . . . . . . . . of course, 10^(-log(7.5)) = 7.5^-1 = 1/7.5
... x ≈ 53,080.96
To find the zeros of this equation you need to first set it equal to zero
2x2 + 16x - 9 = 0
But since you can't FOIL this equation you need to move the non-variable number over
2x2 +16x = 9
Now solve for x by pulling an x out of the equation
x(2x + 16) = 9
x = 9
2x +16 = 9
2x = -7
x = -7/2
So your zeros would be at x = 9, and -7/2
