solve for x in 9^x+32^x-1=53

Question

2)given that p=3y express the equation 3^2y-1 +2x3^y-1=1 terms of p hence or otherwise find the value of y in the equation 3^2y-1+2x3^y-1=1
3)a train travels betwqeen two stations the train starts from the rest and accelerates uniformly for 15 seconds it then travels at aconstant speed for 300 seconds and finally decelerates uniformly for 200 seconds given the distance between two stations is 1045n calculate the:
a)maximum speed in km/h the train attained
b)acceleration
c)distance the train travelling during the last 100 seconds
d)time the train to travel the first half of the journey
4) give me five exrta questions on quadratic equations

Answer 1

1) solve for x:

9^x + 32^x - 1 = 53

Subtract 32^x from both sides:

9^x - 1 = 53 - 32^x

Add 1 to both sides:

9^x = 54 - 32^x

Take the logarithm (any base) of both sides:

x log(9) = log(54 - 32^x)

Divide both sides by log(9):

x = (log(54 - 32^x)) / log(9)

2) Express 3^(2y-1) + 2x3^(y-1) = 1 in terms of p:

3^(2y-1) + 2x3^(y-1) = 1

Replace y with p/3:

3^(2(p/3)-1) + 2x3^((p/3)-1) = 1

Simplify:

3^(2p/3 - 1) + 2x3^(p/3 - 1) = 1

3^((2p-3)/3) + 2x3^((p-3)/3) = 1

Express y in terms of p:

y = p/3

3) Distance between two stations = 1045n

a) Maximum speed:

The train starts from rest and accelerates uniformly for 15 seconds, so we can use the equation:

v = u + at

where u = 0 (starting from rest), a is the acceleration, t is the time.

We don't know the acceleration, but we know the distance and time, so we can use another equation:

s = ut + (1/2)at^2

where s is the distance the train travels during the acceleration phase.

The train accelerates for 15 seconds, so:

s = 0 + (1/2)a(15^2)

s = (225/2)a

The train covers the same distance during the deceleration phase, so we can use the same equation:

s = (225/2)a

The total distance covered by the train is 1045n, so:

1045n = s + 300v + s

1045n = 2s + 300v

Substituting the expressions for s:

1045n = (225/2)a + 300v + (225/2)a

209n = 225a + 300v

We want to find the maximum speed, so we need to know the time when the train stops accelerating and starts traveling at constant speed. This happens after 15 seconds, so the time when the train stops accelerating is 15 + 300 = 315 seconds.

We also know that the acceleration phase and the deceleration phase take the same amount of time (15 + 200 = 215 seconds), so the total time of the journey is:

315 + 215 = 530 seconds

During the constant speed phase, the distance covered by the train is:

1045n - 2s = 1045n - 2(225/2)a = 1045n - 225a

We can use the equation:

v = s/t

where s is the distance covered during the constant speed phase, t is the time.

v = (1045n - 225a)/300

To find the maximum speed, we need to find the maximum value of v. Maximum speed occurs when the acceleration is minimum (or zero), so we can set a = 0:

v = (1045n)/300

Converting to km/h:

v = (1045/300) x (3600/1000) = 12.27 km/h (rounded to two decimal places)

So the maximum speed is 12.27 km/h.

b) Acceleration:

Using the equation:

209n = 225a + 300v

Substituting the values we know:

209n = 225a + 300(12.27)

Solving for a:

a = (209n - 300(12.27))/225

a = (209n - 3671.1)/225

c) Distance traveled during the last 100 seconds:

During the last 100 seconds, the train decelerates uniformly. Let's call the deceleration a'. Using the equation:

v = u + at

where v = 0 (the train comes to a stop), u is the final speed, a' is the deceleration, t is the time.

We can solve for t:

t = -u/a'

We know that the train decelerates for 200 seconds, so the time when it starts decelerating is 315 - 200 = 115 seconds. The time when it stops is 315 + 100 = 415 seconds.

During the constant speed phase (from 115 seconds to 315 seconds), the train covers 300v distance. During the deceleration phase (from 315 seconds to 415 seconds), the train covers:

s' = ut + (1/2)a't^2

where s' is the distance covered during the deceleration phase.

Substituting t:

s' = (u^2)/(2a')

The distance covered during the last 100 seconds is:

s'' = 1045n - 2s - 300v - s'

Substituting expressions for s, s', and v:

s'' = 1045n - 2(225/2)a - 300(12.27) - (u^2)/(2a')

We know that u is the speed the train attained at the end of the constant speed phase, so u = 12.27 km/h.

Substituting u:

s'' = 1045n - 2(225/2)a - 300(12.27) - (12.27^2)/(2a')

We need to find a'. We already found a (the acceleration during the acceleration and deceleration phase). a' is the deceleration during the last 100 seconds. We can use the equation:

v^2 = u^2 + 2as

where v = 0 (the train stops), u = 12.27 km/h, s is the distance the train covers during the last 100 seconds, a' is the deceleration.

Solving for a':

a' = u^2 / (2s)

Substituting u and s:

a' = (12.27^2) / (2s'')

Substituting s'':

s'' = 1045n - 2(225/2)a - 300(12.27) - (12.27^2) / (2[(12.27^2) / (2s'')])

Simplifying:

s'' = 1045n - 225a - 300(12.27) - s''

2s'' = 1045n - 225a - 300(12.27)

s'' = (1045n - 225a - 300(12.27)) / 2

Substituting a:

s'' = (1045n - 225[(209n - 300(12.27))/225] - 300(12.27)) / 2

Simplify:

s'' = (1045n - 209n + 3671.1 - 3670.5) / 2

s'' = 5.3n

So the distance traveled during the last 100 seconds is 5.3n km.

d) Time taken to travel the first half of the journey:

The first half of the journey is the acceleration phase and the first part of the constant speed phase (from 0 to 315/2 = 157.5 seconds). During this time, the train covers:

s = ut + (1/2)at^2

where s is the distance covered during the first half of the journey, u = 0 (starting from rest), a is the acceleration, t is the time.

The time taken to travel the first half of the journey is:

t = s/v

where v is the average speed during the first half of the journey.

The average speed is:

v = 2s / (t_acceleration + t_constant speed)

where t_acceleration is the time taken for acceleration, t_constant speed is the time taken for constant speed.

We know t_acceleration = 15 seconds and t_constant speed = 300 seconds, so:

v = 2s / 315

Substituting the expression for s:

v = 2[(225/2)a + 157.5(12.27)] / 315

Substituting the expression for a:

v = 157.5(12.27) / 315

v = 6.14 km/h

Substituting v into the equation for t:

t = s/v = [(225/2)a + 157.5(12.27)] / 6.14

Substituting the expression for a:

t = [(225/2)[(209n - 300(12.27))/225] + 157.5(12.27)] / 6.14

Simplifying:

t = 313.2n - 238.3

So the time taken to travel the first half of the journey is 313.2n - 238.3 seconds.

4) Five extra questions on quadratic equations:

1) Solve for x: x^2 + 5x + 6 = 0

2) Find the discriminant of the quadratic equation: 3x^2 - 2x + 1 = 0

3) The roots of the equation 2x^2 - 5x + 2 = 0 are a and b. Find the quadratic equation whose roots are a + 1/3 and b + 1/3.

4) Solve for x: 2x^2 + 7x - 15 = 0

5) The difference between the roots of a quadratic equation is 4. If one of the roots is 3, find the equation.

Answer 2

9^x + 32^x - 1 = 53
Your solution of x = (log(54 - 32^x)) / log(9) is bogus, since it has x on both sides.
The solution is x ≈ 1.0861

I suspect that the problem had a typo, and should have been
9^x + 32^(x-1) = 53
because having that -1 on the left is useless.
x ≈ 1.6984

There is no algebraic solution

Answer 3

to 3^(2y-1) + 2x3^(y-1) = 1 in terms of p, because there are no terms involving p in the equation.

For the quadratic equation questions:

1) x^2 + 5x + 6 = 0 can be factored as (x + 2)(x + 3) = 0, so the solutions are x = -2 and x = -3.

2) The discriminant is b^2 - 4ac, so in this case it is (-2)^2 - 4(3)(1) = -8, which is negative. This means the quadratic equation has no real roots.

3) Using Vieta's formulas, we know that the sum of the roots of 2x^2 - 5x + 2 = 0 is -(-5)/2 = 5/2, and their product is (2)/2 = 1. We want to find a quadratic equation whose roots are a + 1/3 and b + 1/3.

(a + 1/3) + (b + 1/3) = a + b + 2/3, which is the sum of the roots of the new quadratic equation.

(a + 1/3)(b + 1/3) = (ab + a/3 + b/3 + 1/9), which is the product of the roots of the new quadratic equation.

We want this new quadratic equation to have the same product (1) and a sum of (5/2 + 2/3), so we can write:

(x - (a + 1/3))(x - (b + 1/3)) = x^2 - [(a + b + 2/3)]x + [(ab + a/3 + b/3 + 1/9)]

Simplifying:

x^2 - [(5/2 + 2/3)]x + [(1)]

x^2 - 17/6 x + 1

So the new quadratic equation is x^2 - (17/6)x + 1 = 0.

4) 2x^2 + 7x - 15 = 0 can be factored as (2x - 3)(x + 5) = 0, so the solutions are x = 3/2 and x = -5.

5) Let the roots be a and b, with a > b. Since the difference between the roots is 4, we know that a - b = 4. We also know that one of the roots is 3, so either a = 3 or b = 3.

If a = 3, then b = -1 (since a - b = 4) and the quadratic equation is (x - 3)(x + 1) = x^2 - 2x - 3 = 0.

If b = 3, then a = 7 (since a - b = 4) and the quadratic equation is (x - 7)(x - 3) = x^2 - 10x + 21 = 0.

So the two possible quadratic equations are x^2 - 2x - 3 = 0 and x^2 - 10x + 21 = 0.