Exponential Equations with Different Bases

Exponential equations with different bases mean we are going to solve the exponential functions with different bases. Normally exponential functions have the base as e. Here we are going to solve the equations other than e. We will see some example problems for exponential functions with different bases. It is better to understand how to work on the different base. If we want to solve the exponential equations we will use the logarithmic values.


Examples for Exponential Equations with Different Bases:

Solve the following equation 3x = 275x + 3

Solution:

The given equation is 3x = 275x + 3

So we get 3x = ((3)3)5x + 3

From this 3x = (3)15x + 9

From the above we can get x = 15x + 9

15x – x = -9

14x = -9

x = `-9 / 14`

Example 2 for exponential equations with different bases:

Solve the following equation 2x = 42x + 4

Solution:

The given equation is 2x = 42x + 4

So we get 2x = ((2)2)2x + 4

From this 2x = (2)4x + 8

From the above we can get x = 4x + 8

4x – x = -8

3x = -8

So x = `-8 / 3`
More Examples for Exponential Equations with Different Bases:

Example 3 for exponential equations with different bases:

Solve the following equation 5x = 125x + 3

Solution:

The given equation is 5x = 125x + 3

So we get 5x = ((5)3) x + 3

From this 5x = (5)3x + 9

From the above we can get x = 3x + 9

3x – x = -9

2x = -9

So x = `-9 / 2`

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Example 4 for exponential equations with different bases:

Solve the following equation 4x = 16x + 4

Solution:

The given equation is 4x = 16x + 4

So we get 4x = ((4)2) x + 4

From this 4x = (4)2x + 4

From the above we can get x = 2x + 4

x – 2x = 4

-x = 4

So x = 4

Logarithmic Expression Calculator

In mathematics, the logarithmic expression of an integer with a provided base is the power or exponent to which the base must be increased in order to produce the integer. For example, the logarithmic function of 1000x to base 10 is 3x, because 3 is the power to which ten must be raised to produce 1000: 103 = 1000, so log101000x  = 3x. Only positive real integers will have the real integer values; negative and complex integers hold complex logarithms.Now we are going to discuss about logarithmic expression calculator in detail.



Logarithmic Expression Calculator:

Logarithmic expression calculator is used to calculate the expression given in terms of logarithmic form. Logarithmic calculator calculates the values for the logarithmic problems. For every logarithmic function there is a value present in logarithm form. sample lograthmic calculator is given below. Here the calculate button is used to find the value of the given question. Reset button is used to clear the values and the question present in the calculator, which is useful to enter the next problem. sample calculator is given below,

Logarithmic Expression Calculator

Rules of lograthmic expression solving.

If ax = N, then loga N = x

logx mn = logx m + logx n

logx (m/n) = logx m - logx n

logx mn = n logx m

logn m = logx m/ logx n

logn m * logo n * logp o = logp m

logn m * logm n = 1, logm m = 1

logn m = `1/ (log_m n)`

logx 1 = 0 . Log 1 to any base is always 0.

these are the formulas for expression using in logarithmic expression calculator.
Exampl for Logarithmic Expression Calculator:

Exampl for logarithmic expression calculator 1:

Solve logarithmic function log4(9x)5

Solution:

Step 1: The given function is log4(9x)5

Step 2: To solve this function using log function rules

Rule 3: logb (mn) = n· logb (m)

Step 3: log4 (9x)5= 5 log4 (9x)

So the solution is log4(9x)5 = 5 log4 (9x)

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Exampl for logarithmic expression calculator 2:

Solve logarithmic function log3 `((99/2)x)`

Solution:

Step 1: The given function is log3`((99/2)x)`

Step 2: To solve this function by help of using log function rules

Rule 1:  logb (mn) = logb (m) + logb (n)

Step 3: log3 `((99/2)x)` = log3 `(99/2)`+ log3(x)

Rule 2:  logb (m/n) = logb (m) - logb (n)

log `(99/2)` = log (99) - log (2)

So the solution is log3  `((99/2)x)` = log3 (99) - log3 (2) + log3(x)

Exampl for logarithmic expression calculator 3:

Solve logarithmic functionlog2 (99/x)

Solution:

Step 1: The given function is log2 (99/x)

Step 2: To solve this function by help of using log function rules

Rule 2: logb (m/n)) = logb (m) – logb (n)

Step 3:  log2 (99/x) = log2 (99) - log2(x)

So the solution is log2 (99/x) = log2 (99) - log2(x)

Solving Simple Differential Equation

Solving simple differential equations involve the process of differentiating the algebraic function with respect to the input function. The algebraic function which is differentiable is known as differential equations. The differential equation comes under calculus category whereas to find the rate of change of the given function with respect to the input function. The following are simple example differential equations for solving.


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Simple Differential Equations Examples for Solving:

The following are the example problems with simple differential equations for solving.

Example 1:

Solve the simple differential equation.

f(k) = k2 – 4k + 8

Solution:

The given equation is

f(k) = k 2 – 4k + 8

The first derivative f ' for the algebraic function is

f '(k) = 2 k  – 4

Example 2:

Solve the simple differential equation.

f(k) = k 3 – 5 k 2  + 11k

Solution:

The given function is

f(k) = k 3 – 5 k 2  + 11k

The first derivative f ' for the algebraic function is

f '(k) = 3k 2 – 5(2 k  ) + 11

f '(k) = 3k 2 – 10 k + 11

Example 3:

Solve the simple differential equation.

f(k) = k4 – 3k 3 – 4k 2  + k

Solution:

The given function is

f(k) = k4 – 3k 3 – 4 k 2  + k

The first derivative f ' for the algebraic function is

f '(k) = 4 k 3 – 3(3k 2 ) – 4( 2 k  ) + 1

f '(k) = 4 k 3 – 9k 2  – 8 k  + 1

My forthcoming post is on Example of Hypothesis Testing and Dividing Fraction will give you more understanding about Algebra.

Example 4:

Solve the simple differential equation.

f(k) = k 5 – 6 k 3  + 11

Solution:

The given function is

f(k) = k 5 – 6 k 3  + 10

The first derivative f ' for the algebraic function is

f '(k) = 5k 4 – 6(3 k 2 )

f '(k) = 5k 4 – 18 k 2
Simple Differential Practice Equations for Solving:

1) Solve the simple differential equation.

f(k) = k 3 – 6 k 2  + 11k

Answer: f '(k) = 3k 2 – 12 k

2) Solve the simple differential equation.

f(k) = k 2 – 6 k   + 11

Answer: f '(k) = 2k – 6

About Metric Volume Units

The quantities used to find area, length, width capacities and volume of things etc are called measures. Many countries have their own system of measures. But Metric System of measures is very simple and easy to calculate. The area is measured in square unit. In metric system the volume is measured in cubic units.
Example Problems - Metric Volume Units:

The triangular prism has width 6 cm, height 9 cm and length 11 cm. find the volume of triangular prism.

Solution:

Given:

Width (w) = 6 cm

Height (h) = 9 cm

Length (l) = 11 cm

Formula:

Volume of triangular prism (V) = `1/2` (l x w x h) cubic units

= `1/2` (11 x 6 x 9)

= `1/2` (594)

= 297

Volume of triangular prism (V) = 297 cm3

2. figure out the volume trapezoidal prism whose length 11 cm, height 8cm, length of parallel sides a=7 cm and b=4cm.

Solution:

Given:

Length (l) = 11cm

Height (h) = 8cm

Parallel sides a=7cm and b=4cm

Formula:

Volume of trapezoidal prism = l x area of the base cubic units

Area of the base:

Area of the base = `1/2` x (a + b) x h

= `1/2` x (7 + 4) x 8

=`1/2` x 11 x 8

= 44 cm2

Volume of trapezoidal prism = 11x 44

= 484

Volume of trapezoidal prism = 484 cm3

3. The cylinder has the radius r = 10 feet, h=23 feet. Find the volume of cylinder.

Solution:

Given:

r=10 cm

h=23 cm

Formula:

The volume of the cylinder = π x r2 x h cubic unit

=3.14 x (10)2 x 23

The volume of the cylinder = 7222 ft3.
Example Problems - Metric Volume Units:

Cone:

4. The cone has the radius = 10 feet and height = 23 feet. Find the volume of the cone.

Solution:

Given:

Radius (r) = 10 feet

Height (h) = 23 feet

Formula:

The volume of the cone =`1/3` x π x r2  x  h

= `1/3` x 3.14 x (10)2 x 23

The volume of the cone = 2407.33 ft3

5. What is the volume rectangular prism with length 8 cm width 5 cm and height 6 cm?

Solution:

Given:

Length =8 cm

Width = 5 cm

Height = 6 cm

Formula:

Volume of rectangular solid (v) = l x w x h

= 8 x 5 x 6

= 240

Volume of rectangular solid (v) = 240 cm3

radius calculation formula

Pi is also one of the important concepts in algebra. The value of pi is 22/7 or 3.14. Using these values we can easily find the answer for complex problems in algebra. We have different kinds of problems using pi in algebra. Commonly the pi formulas are used to find area or volume of the specific shape in algebra. Here we are going to learn some problems for calculate radius.
Radius Calculation Formulas:

Perimeter of a circle = 2 * pi * radius.
Area of circle = Pi * radius 2.
Area of ellipse = pi *radius1 *radius 2.
Surface area of sphere = 4*pi* radius 2.
Surface area of cylinder = 2 * pi *radius *height.
Surface area of cone: pi* radius* side.
Surface area of torus = pi2 * (radius2 2 –radius12).
Volume of sphere = `(4)/(3)` *pi*radius 3.
Volume of ellipsoid = `(4)/(3)` *pi*radius 1*radius 2*radius3.
Volume of cylinder =pi*radius*height.
Volume of cone = `(1)/(3)` * pi* radius 2 *height.
Volume of torus = `(1)/(4)` * pi2 *(radius1 + radius2) *(radius1-radius2)2
Volume of hemisphere = `(2)/(3)` pi r3.

These are the formulas to calculate radius.
Example Problems of Radius Calculation Formula:

Example 1:

Perimeter of a circle is 40 m .calculate the radius of the circle?

Solution:

Step 1: Perimeter of a circle = 2 * pi * radius.

Step 2: We know that perimeter = 40 m.

Step 3: Plug the perimeter value in to the formula.

Step 4: 40 = 2 * 3.14 * radius.

Step 5: 40 = 6.28 *radius. (Divide using 6.28 on both the sides)

Step 6: Therefore, radius = 6.4 m.

Example 2:

Surface area of cone is 120 cm^2 and side = 7cm.Calculate the radius of the cone?

Solution:

Step 1: Surface area of cone: pi* radius* side.

Step 2: We know that surface area and side.

Step 3: Plug the surface area and side value in to the formula.

Step 4: Therefore, surface area of cone is 120 = 3.14 * radius *7.

Step 5: So, 120 = 21.98 *radius. (Divide using 21.98 on both the sides)

Step 5: Therefore radius = 5.5 cm.

These are the example problems of calculate radius using formula.

Practice problems of calculate radius using formula.

1)   Volume of cylinder = 160 m and height = 9 cm .calculate the radius of the cylinder?

2)   Suppose the area of circle is 210 m. calculate the radius of the circle?

Answer key

1)   Radius = 5.7 m

2)   Radius = 8.17 m

Dividing Decimal Word Problems

Decimal is one of the number systems which have the base value 10. Division is one of the basic arithmetic operations which is the inverse of multiplication operation. Dividing decimal word problems occurs in such circumstances where we are trying to find out how many times a decimal number go into another. In this article we will see few examples of dividing decimal word problems.

Dividing Decimal Word Problems Examples:

Example 1:

If 2.5 kg of apples cost `$` 15.50 what is the price of one kilogram of apple?

Solution:

Here to find the cost of 1kg of apple we need to divide 15.50 by 2.5

Division of 15.50 by 2.5

By dividing `15.50/2.5` = 6.2

Hence the price of 1 kg apple = `$` 6.2

Example 2:

Tom wants to buy chocolates. The cost of one chocolate is `$` 0.5. The total amount tom has is `$` 20.25. How many chocolates he can buy?

Solution:

Total amount Tom has =` $` 20.25

Cost of one chocolate = `$` 0.5

Number of chocolate can buy =` 20.25 / 0.5`

Division of 20.25 by 0.5

= 40.5

Hence Tom can buy 40 chocolates for `$` 20.25

Example 3:

Each bag of cherries weighs 2.25 pounds. How many bags of cherries would be needed to make 92.25 pounds?

Solution:

Weight of 1 bag of cherries = 2.25 pounds

Total pounds = 92.25

To find number of bags of cherries,

Divide 92.25 by 2.25

Division of 92.25 by 2.5

By dividing` 92.25 /2.25` = 41

Hence 41 bags of cherries needed to make 92.25 pounds.


Dividing Decimal Word Problems Examples Continued:

Example 4:

There are 15 students in a class.  Together they weigh 1925.90 pounds.  Find the average weight?

Solution:

Total Number of students = 15

Total weight = 1925.90 pounds

Average weight = `1925.90/15`

Division of 1925.90 by15

=128.39pounds

Example 5:

There are 2.54 centimeters in one inch.  How many inches are there in 75.24 centimeters?

Solution:

1 inch = 2.54cm

75.24cm= `75.24/2.54`

Division of 75.24 by 2.54

= 29.62inches

Hence there are 29.62inches in 75.24cm

Example 6:

The scores of three persons in diving competition are 3.4, 5.75, and 6.61.  What is the average score?

Solution:

Average score = Total score/ number of persons

Total score =   3.4+5.75+6.61=15.76

Average score = `15.76/3`

Division of 15.76 by3

= 5.25

Two Equations in Standard Form

Linear equation is an algebraic equation which has constants and variables together. Linear equation has 1 or more variables. Linear equation has more forms. Standard form is one of the form of linear equation.

Standard form of linear equation is Ax + By = C.

Here A, B and C are constants. X and y are variables. A and B are not zero.

We can solve standard form of equation using substitution or elimination method. But here we use two standard form equations. Let us see how to solve.

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Standard Form of Two Equations – Substitution Method:

Problem:

Solve these two standard forms of equation using substitution method.

2x + 4y – 12 = 0

X – 2y – 4 = 0

Solution:

The given standard form of two equations are,

2x + 4y = 12  (1)

X – 2y = 4  (2)

From (2), we rewrite the equation as

X = 4 + 2y  (3)

Substitute equation (3) into (1) to find the variable y.

2(4+2y) + 4y = 12

Apply the distributive property, we get

8 + 4y + 4y = 12

Combine like terms,

8+ 8y = 12

Subtract 8 from each side.

8 – 8 + 8y = 12 – 8

8y = 4

Divide by 8 each side.

`(8y)/8` = `4/8`

Y = `1/2`

Substitute y = `1/2` into equation (2)

X – 2y = 4

X – 2(`1/2` ) = 4

X – 1 = 4

Add 1 to each side.

X – 1 + 1 = 4 + 1

X = 5.

Therefore, the solutions are 5 and 1/2.

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Standard Form of Two Equations – Elimination Method:

Problem :

Use elimination method to determine the solutions of the following the systems of equations.

x + y – 16 = 0 and 4x – 2y – 4 = 0

Solution:

The given standard forms of two equations are

x + y = 16  (1)

4x – 2y = 4  (2)

Step 1:

Multiply the equation (1) by 2 and Equation (2) by 1 to get the coefficients of variable y same. So the equations are,

2x + 2y = 32

4x – 2y = 4

Step 2:

Add the two equations for eliminating y variable.

2x + 2y + 4x – 2y = 32 + 4

2x + 4x + 2y – 2y = 36

6x = 36

Divide by 6 both sides.

x = 6

Step 3:

Substitute the x value into the equation (1) to get value of y variable.

x + y = 16

6 + y = 16

Subtract 6 from each side.

y = 10.

The solutions are x = 6 and y = 10.