solve these questions: 8^x-1 = 4, x^/2 = 2 1/4 and (1/2)^x = (1/64)^2x+3
as well as:
8^x-1 = 4
x^/2 = 2 1/4
(1/2)^x = (1/64)^2x+3
Solution
To solve the first exponential equation, we\'ll have to create matching bases, such as;
8^(x-1) = (2^3)^(x-1)
We\'ll multiply the superscripts:
8^(x-1) = 2^3(x-1)
Now, we\'ll manage the right side and we\'ll write 4 as a power of 2:
4 = 2^2
We\'ll re-write the equation:
2^3(x-1) = 2^2
Since the bases are matching, we\'ll apply one to one rule:
3(x-1) = 2
3x - 3 = 2
3x = 3 + 2
3x = 5
x = 5/3
The solution of the 1st. equation is x = 5/3.
Since the 2nd expression is not so clear, we\'ll solve the 3rd equation.
We\'ll manage the right side and we\'ll re-write 1/64 as a power of 1/2.
1/64 = (1/2)^6
We\'ll raise both sides to the power (2x+3):
(1/64)^(2x+3) = (1/2)^6(2x+3)
We\'ll re-write the equation:
(1/2)^x = (1/2)^6(2x+3)
Since the bases are matching, we\'ll apply one to one rule:
x = 6(2x+3)
x = 12x + 18
11x = -18
x = -18/11
The solution of the 1st equation is x = 5/3 and the solution of the 3rd equation is x = -18/11..
solve these questions with detail stepsThe X-hose is a commercial .pdf
1. solve these questions with detail steps
The X-hose is a commercial elastic garden hose, which expands during use. You can find
advertising material and a movie of its use here: https://www.xhose.com/
The hose comes with a standard end nozzle/valve. The patent application suggests that the nozzle
is required to ensure a high pressure inside the hose. i) Are they correct? Provide calculations to
support your arguments. ii) Provide an estimate of the flow rate from the hose. Household water
pressure (Head) in Sydney is about 20meters. The expanded hose is about 30meters long and has
a diameter of 15mm. Assume all other quantities and apply your engineering judgement to
simplify the problem.
Solution
these are correct,,because for engineering point of view there losses are very low and the losses
can be calculated by formula flv^2/2dg..if losses are low then the hose will work with full
pressure,,and the flow rate can be calculated by using Q=AREA*VELOCITY..where area is
pi*d*l..
thank you if any doubt then please ask
