You can take the log of the left and right hand side, and then apply the <span>logarithm rules:
log(a</span>ˣ) = x·log(a)
log(ab) = log(a) + log(b)
log(9^(x-1) * 2^(2x+2)) = log(6^(3x))
log(9^(x-1)) + log(2^(2x+2)) = 3x log(6)
(x-1) log(9) + (2x+2) log(2) - 3x log(6) = 0
x(log9 + 2log2 - 3log6) = log9 - 2log2
x = (log9 - 2log2) / (log9 + 2log2 - 3log6)
simplifying by writing log9 = 2log3 and log6 = log2+log3
x= 2(log3 - log2) / (2log3 + 2log2 - 3log2 - 3log3) =
x= -2(log3 - log2) / (log3 + log2) = -2 log(3/2) / log(6)
So 6^x = 4/9
Since we are given the values for both rows only in the second column, we can use this to solve for the rest of the missing values. Simply divide 3 by 2.49 and multiply that resultant by and multiply that by any missing value for cookies to find the missing cost. In order to solve for the cookies when cost is given, divide 2.49 by 3 and multiply that resultant. I will solve for the first, third, and fourth columns.
3/29=1.2
For column 1, 1.2*1=1.20
For column 3, 1.2*20=24.10
For column 4, 1.2*100=12.00
Answer:
There are 10 slices left
Step-by-step explanation:
2x^2 +3x -4 +8 -3x -5x^2 +2
-3x^2 + 6
Answer is C) -3x^2 + 6
