The answer is log3 k to the seventh power m to the sixth power over n to the ninth power
a * logₓ(y) = logₓ(yᵃ)
7 log₃ (k) = log₃ (k⁷)
6 log₃ (m) = log₃ (m⁶)
9 log₃ (n) = log₃ (n⁹)
7 log₃ (k) + 6 log₃ (m) - 9 log₃ (n) = log₃ (k⁷) + log₃ (m⁶) - log₃ (n⁹)
logₓ(y) + logₓ(z) = logₓ(y * z)
log₃ (k⁷) + log₃ (m⁶) - log₃ (n⁹) = log₃ (k⁷ * m⁶) - log₃ (n⁹)
logₓ(y) - logₓ(z) = logₓ(y / z)
log₃ (k⁷ * m⁶) - log₃ (n⁹) = log₃ (k⁷ * m⁶ / n⁹)
It would round up to 1,244,000
The answer is 3. I hope this helps!
X-intercept: (3.5,0)
Y-intercept: (2.5,0)
Answer:
The number of hours Jackie worked = 80hours
Step-by-step explanation:
Last summer:
Jackie made $960
Rachel made $960
let number of hours Jackie worked = x
Rachel worked 16 hours more than Jackie:
Number of hours Rachel worked = x + 16
if Jackie earned $y per hour
Rachel earned $2 less per hour = y-2
Jackie: 960 = x × y = xy
Rachel: 960 = (x+16)(y-2)
960 = xy -2x +16y -32
recall xy = 960, insert the value for xy
960 = 960 - 2x +16y -32
- 2x +16y -32 = 0
2x -16y = -32
x-8y = -16
x = 8y-16
recall xy = 960, insert the expression for x
(8y-16)y = 960
8y² -16y = 960
y² -2y - 120 = 0
y²+10y-12y -120 = 0
y(y+10) -12(y+10) = 0
(y-12) = 0 or (y+10) = 0
y = 12 or -10
since y can't be negative, y = 12
x = 8y-16
x = 8(12) -16 = 80
The number of hours Jackie worked = x = 80 hours
