Solve: log5x^3 - logx^2 = 2

x = 1/5 exp((2 - 12 log(5) + 2^(1/3) (56 - 18 log(5) + 27 log^2(5) + 3 √(348 - 216 log(5) + 324 log^2(5) - 12 log^3(5) + 81 log^4(5)))^(2/3))/(3 2^(2/3) (56 - 18 log(5) + 27 log^2(5) + 3 √(348 - 216 log(5) + 324 log^2(5) - 12 log^3(5) + 81 log^4(5)))^(1/3)) + 1/3) or x = 1/5 exp(1/3 - (2 - 12 log(5) + 2^(1/3) (-56 + 18 log(5) - 27 log^2(5) - 3 √(348 - 216 log(5) + 324 log^2(5) - 12 log^3(5) + 81 log^4(5)))^(2/3))/(3 2^(2/3) (-56 + 18 log(5) - 27 log^2(5) - 3 √(348 - 216 log(5) + 324 log^2(5) - 12 log^3(5) + 81 log^4(5)))^(1/3))) or x = 1/5 exp((-2 (-2)^(1/3) + 12 (-2)^(1/3) log(5) + (-112 + 36 log(5) - 54 log(5)^2 - 6 √(348 - 216 log(5) + 324 log(5)^2 - 12 log(5)^3 + 81 log(5)^4))^(2/3))/(6 (56 - 18 log(5) + 27 log^2(5) + 3 √(348 - 216 log(5) + 324 log^2(5) - 12 log^3(5) + 81 log^4(5)))^(1/3)) + 1/3)

Step-by-step explanation:

see attachment

b. 1.25 hr

step-by-step explanation:

if you have a constant speed of 60 mph, then you have a ratio between 60 mi. to 1 hr. dividing both sides by 60 we get the ratio

1 mi. to

then mult. both sides by 75 we get the ratio

75 mi. to

simplifying we get

75 mi. to

or 1.25 hr.

b, i believe, but not sure.