Archive for October, 2015

Problem 1115

Solve the logarithmic equation. Be sure to reject any value of x that is not the domain of the original logarithmic expression.

6 In(8x) = 12

What is the exact solution?

What is the decimal approximation to the solution?

 

Solution:-

 

The solution set is {\frac{e^{2}}{8}}.

The decimal solution set is {0.92}.

 

 

Problem 1114

Solve the following logarithmic equation. Be sure to reject any value of x that is not in the domain if the original logarithmic expression. Give the exact answer.

{log_{2}}^{3x+3} = 4

 

Solution:-

 

The solution set is {\frac{13}{3}}.

 

 

Problem 1113

Solve the following logarithmic equation. Be sure to reject any value of x that is not in the domain of the original logarithmic expression. Give the exact answer.

{log_{4}}^{x}.

 

Solution:-

 

The solution set is {64}

 

 

Problem 1112

What point is symmetric with respect to the y-axis to the point (\frac{\sqrt{3}}{2},-\frac{1}{2}) ?

 

Solution:-

 

Recall for any point (x, y) the point symmetric about the y-axis to (x, y) is the point (-x , y).

If two point are symmetric about the y-axis, then the points have the same y-coordinate and opposite x-coordinates. Replace the x-coordinate of the given point with its opposite.

Identify the x-coordinate of the given point, (\frac{\sqrt{3}}{2},-\frac{1}{2}).

x = \frac{\sqrt{3}}{2}

If x = \frac{\sqrt{3}}{2},then the opposite of x, -x = – \frac{\sqrt{3}}{2}

Therefore , the point symmetric to the point (\frac{\sqrt{3}}{2},-\frac{1}{2}) with respect to the y-axis is (-\frac{\sqrt{3}}{2},-\frac{1}{2}).

 

 

 

Problem 1111

Find the central angle θ which forms a sector of area 21 square feet of a circle of radius 11 feet.

 

Solution:-

 

A central angle is an angle whose vertex is at the center of a circle. The rays of a central angle subtend (intersect) an arc on the circle. The part of the circle between the rays of the angle and the arc subtended is called a sector.

Area of a sector

The area A of sector of a circle of radius r formed by a central angle of θ radians is equal to the following

A = \frac{1}{2} r2 θ

Solving for θ given θ = \frac{2A}{r^{2}}.

Substitute the values for A and r, and simplify.

θ =\frac{2A}{r^{2}}

=\frac{2 \cdot 21}{11}

\approx 0.347

Therefore, the central angle θ which forms a sector of area 21 square feet of a circle of radius 11 feet is θ = 0.347 radians.

 

 

 

Problem 1110

Find the area A of the sector of a radius 30 feet formed by the central angle \frac{1}{13} radian.

 

Solution:-

 

A central angle is an angle whose vertex is at the center of a circle. The rays of a central angle subtend (intersect) an arc on the circle. The part of the circle between the rays of the angle and the arc subtended is called a sector. We have the following theorem.

Area of a sector

The area A of sector of a circle of radius r formed by a central angle of θ radians is

A = \frac{1}{2} r2 θ

The value of r in the formula is 30 feet.

The value of θ in the formula is \frac{1}{13} radian.

Therefore,

A (area) = \frac{1}{2}*(30 feet)2 * \frac{1}{13}

= 34.615 feet2

 

 

Problem 1109

Find the length s of the arc of a circle of radius 85 centimeters subtended by the central angle 36°.

 

Solution:-

 

A central angle is an angle whose vertex is at the center of a circle. The rays of a central angle subtend (intersect) an arc on the circle. We have the following theorem.

Arc Length

For a circle of radius r, a central angle θ radians subtends an arc whose length s is

s = r θ

Convert angle in degrees to radians.

1° = \frac{\pi}{180} radian

36° = 36 *\frac{\pi}{180} radian

\approx \frac{\pi}{5}radian

s(arc length) = rθ

= 85 centimeters* \frac{\pi}{5}

=17π centimeters

\approx 53.407 centimeters

 

 

Problem 1108

Find the central angle θ which subtends an arc of length 53 miles of a circle of radius 51 miles.

 

Solution:-

 

A central angle is an angle whose vertex is at the center of a circle. The rays of a central of a circle. The rays of a central angle subtend (intersect) an arc on the circle. The arc length for a circle of radius r and a central angle of θ radians is given by s = r θ.

Solving this expression for θ given θ = \frac{s}{r}. Substitute the values for s and r.

θ = \frac{s}{r}

=\frac{53}{51}

\approx 1.039 radians

Therefore, the central angle which subtends an arc of length 53 miles of a circle of radius 51 miles is θ = 1.039 radians.

 

 

Problem 1107

Find the length s of the arc of a circle of radius 75 meters subtended by the central angle \frac{1}{25} radian.

 

Solution:-

 

A central angle is an angle whose vertex is at the center of a circle. The rays of a central angle subtend (intersect) an arc on the circle. We have the following theorem.

Arc Length

For a circle of radius r, a central angle of θ radians subtends an arc whose length s is

s = r θ

the value of r in the formula is 75 meters.

The value of θ in the formula is \frac{1}{25} radian.

s(arc length) = 75 meters * {1}{15} radian

= 3 meters

 

 

Problem 1106

Convert the angle in radians to degrees. Express your answer in decimal form.

5.63

 

Solution:-

 

Consider a circle of radius r. A central angle of a revolution (360°) will subtend an arc equal to the circumference of the circle. Because the circumference of the circle equal 2πr, we see that 360° = 2π radians, or 180° = π radians. Dividing both sides by π, we have the following conversion formula.

Converting form radians to degrees can be summarized as follows.

1 radian = \frac{180}{\pi} degrees

Therefore, the following is true.

5.63 radians = 5.63 * 1 radian

= 5.63 *\frac{180}{\pi} degrees

\approx 322.58°