Category Archives: Algebra

5.4 Exponential and Logarithmic Equations, 6.6 Logarithmic and Exponential Equations, Exponential Equations

Solve an exponential equation: Take the log of both sides

Example:

(1.41)^x = (sqrt{2})^{1-4x}

Solution:

 

The exponential equation
 (1.41)^x = (sqrt{2})^{1-4x}
Since the bases cannot be easily written the same use the method of taking the log of both sides
(1.41)^x = (sqrt{2})^{1-4x}
ln (1.41)^x = ln (sqrt{2})^{1-4x}
Use the power rule for logarithms
ln (1.41)^x = ln (sqrt{2})^{1-4x}
x ln (1.41) = (1-4x) ln sqrt{2}
Use the distributive law
x ln (1.41) = (1-4x) ln sqrt{2}
x ln (1.41) = ln sqrt{2}-4x ln sqrt{2}
Collect the terms with x to one side and collect the terms without x on the other side
x ln (1.41) = ln sqrt{2}-4x ln sqrt{2}
x ln (1.41) +4x ln sqrt{2}= ln sqrt{2}-4x ln sqrt{2} +4x ln sqrt{2}
x ln (1.41) +4x ln sqrt{2}= ln sqrt{2}
Factor the common x
x ln (1.41) +4x ln sqrt{2}= ln sqrt{2}
x( ln (1.41) +4 ln sqrt{2})= ln sqrt{2}
Solve for x by dividing both sides by the factor in the parenthesis and simplify
x( ln (1.41) +4 ln sqrt{2})= ln sqrt{2}
{x( ln (1.41) +4 ln sqrt{2})}/{ ln (1.41) +4 ln sqrt{2}}= {ln sqrt{2}}/{ ln (1.41) +4 ln sqrt{2}}
x = {ln sqrt{2}}/{ ln (1.41) +4 ln sqrt{2}}
The solution
x = {ln sqrt{2}}/{ ln (1.41) +4 ln sqrt{2}}
x =0.2003

When you type this into a calculator be sure to use parenthesis around the numerator and around the denominator.  Here is an example of how you might enter it.

(ln (sqrt{2}))/(ln(1.41)+4 ln(sqrt{2}))

Here is a youtube video with a similar example.

5.3 Properties of Logarithms, 5.4 Exponential and Logarithmic Equations, 6.6 Logarithmic and Exponential Equations, Logarithmic Equations

Solve the Logarithmic Equation by the one to one property

Example:

2 log_3(7-x)-log_3 2=log_3 18

Solution:

 

The logarithmic equation
2 log_3(7-x)-log_3 2=log_3 18
Use the power rule and the quotient rule to condense to a single logarithm
2 log_3(7-x)-log_3 2=log_3 18
log_3(7-x)^2-log_3 2=log_3 18
log_3((7-x)^2/ 2)=log_3 18
Since both sides of the equation have the same log base the expressions inside the logarithms must be equal
log_3((7-x)^2/ 2)=log_3 18
(7-x)^2/ 2= 18
Clear the denominator by multiplying by 2 on both sides and simplifying
(7-x)^2/ 2= 18
2*(7-x)^2/ 2= 2*18
(7-x)^2= 36
Get rid of the square by square rooting both sides and simplifying
(7-x)^2= 36
sqrt{(7-x)^2}= sqrt{36}
7-x= pm 6
Get x by itself by subtracting 7 on both sides
7-x= pm 6
7-7-x=-7 pm 6
-x=-7 pm 6
Get x by itself by dividing both sides by negative 1
-x=-7 pm 6
-x/-1={-7 pm 6}/-1
x=7 pm 6
x=7 + 6 or x=7 - 6
x=13 or x=1
Check x=13
2 log_3(7-13)-log_3 2=log_3 18
2 log_3(-6)-log_3 2=log_3 18
Log of a negative is undefined.  Exclude this solution.
Check x=1
2 log_3(7-1)-log_3 2=log_3 18
2 log_3(6)-log_3 2=log_3 18
log_3(6)^2-log_3 2=log_3 18
log_3 36-log_3 2=log_3 18
log_3 36/2=log_3 18
log_3 18=log_3 18
Keep this solution.

The solution to the equation is x=1.

Here is a youtube video that is similar.

5.1 Exponential Functions, 6.6 Logarithmic and Exponential Equations, Exponential Equations

Solving an Exponential Equation: Relating the Bases

Example:  Solve the exponential equation.

4^{x-9}=1/1024

Solution:

The exponential equation
4^{x-9}=1/1024
Try to write both sides of the equation with the same base.  Try 4 since there is a base of 4 on the left 4^{x-9}=1/4^5
Using a property of negative exponents move the base to the numerator  4^{x-9}=4^{-5}
Now that that the bases are the same the exponents must be equal  4^{x-9}=4^{-5}

x-9=-5
Solve for x
  x-9=-5

x-9+9=-5+9

x=4

The solution the the exponential equation is 4.

Here is a youtube video with a similar example.

 

3.1 Functions, 3.1 Relations and Functions, Difference Quotient

Difference Quotient: Quadratic Function

Example:  Find the difference quotient for f(x)=x^2-9x

The Difference Quotient:{f(x+h)-f(x)}/h

Solution:

The Difference Quotient Formula
{f(x+h)-f(x)}/h
Write the difference quotient for the given function
={(x+h)^2-9(x+h)-(x^2-9x)}/h
Apply the exponent and use the distributive property
={(x+h)(x+h)-9x-9h-x^2+9x}/h
Multiply
={x^2+xh+xh+h^2-9x-9h-x^2+9x}/h
Combine the like terms
={x^2+2xh+h^2-9x-9h-x^2+9x}/h
Combine the like terms.  Only terms with h should remain
={2xh+h^2-9h}/h
Divide h into each term
={2xh}/h+{h^2}/h-{9h}/h
Cancel the common h from each term
=2x+h-9

1.2 Applications of Linear Equations, Applications of Linear Equations

Application of Linear Equation (Simple Interest)

Example:  Larry invested part of his $31,000 advance at 6% annual simple interest and the rest at 5% annual simple interest.  If his total yearly interest from both accounts was $1,760, find the amount invested at each rate.

Solution:  This question involved simple interest.  We will use the simple interest formula.

I=Prt

The I stands for interest, the P stands for principal (initial investment), r stands for the interest rate and t stands for time in years.

When starting an application problem like this it can be helpful to organize the information in a table using the formula.

I P r t
6% Account
5% Account

Begin filling in the information from the problem.  The interest rate is the most obvious and you should use the decimal version of the percent by moving the decimal place left two places.  Since the problem state that $1760 is the yearly interest for both accounts, we will use a time of 1 year.

I P r t
6% Account .06 1
5% Account .05 1

The goal in this question is to find the amount invested in each account.  I will let x be the amount invested in the 6% account.  The problem states that the rest of the $31,000 will be invested in the 5% account.  We can represent the rest of the money as 31000-x.

I P r t
6% Account x .06 1
5% Account  31000-x .05 1

You can fill in the interest column using the simple interest formula.

For the 6% account: I=Prt=(0.06)(x)(1)=0.06x

For the 5% account: I=Prt=(0.05)(31000-x)(1)=0.05(31000-x)

I P r t
6% Account 0 .06x x 0.06 1
5% Account 0.05(31000-x) 31000-x 0.05 1

You can now create an equation with this information.  The interest from the first account plus the interest from the second account should equal the total interest of $1760.

0.06x+0.05(31000-x)=1760

Solve the equation.
 0.06x+0.05(31000-x)=1760
Use the distributive property and combine like terms to simplify each side of the equation.
0.06x+1550-0.05x=1760
0.01x+1550=1760
Solve for x by subtracting 1550 on both sides and simplifying.
0.01x+1550-1550=1760-1550
0.01x=210
Solve for x by dividing both sides by 0.01 and simplifying.
{0.01x}/0.01=210/0.01
x=21000

Since x represents the amount invested in the 6% account, $21000 is invested in the 6% account and the rest is invested in the 5% account.  The rest is $31,000-$21,000=$10,000.  $10,000 is invested in the 5% account.

Video Example:

1.1 Linear Equations, Solving Equations

Rational Equation (no solution)

Example: Solve the rational equation.

{x+4}/{x-3}+1=7/{x-3}

Solution:

{x+4}/{x-3}+1=7/{x-3}

Since we are solving a rational equation we need to first find the restrictions (the values of x that cause the expression to be undefined).

To find the restrictions create an equation by setting each denominator equal to zero and solving.

x-3=0

x-3+3=0+3

x=3

Having x=3 causes a zero in the denominator and the overall expression undefined.  That makes 3 a restricted value .

With the restriction in mind we will solve the equation.

 

The original equation
 {x+4}/{x-3}+1=7/{x-3}
Multiply each side of the equation by the least common multiple of the denominators.  For this equation the least common multiple is x-3
(x-3)({x+4}/{x-3}+1)=(x-3)7/{x-3}
Distribute the least common multiple to each term.
(x-3){x+4}/{x-3}+1(x-3)=(x-3)7/{x-3}
Simplify by canceling the common factors.  This should clear any denominators.
x+4+1(x-3)=7
Use the distributive property to simplify.
x+4+x-3=7
Simplify each side of the equation by combining like terms.
2x+1=7
Solve for x by getting x by itself on one side.  Start by subtracting 1 on both sides.
2x+1-1=7-1
2x=6
 Solve for x by getting x by itself on one side.  Next divide both sides by 2.
{2x}/2=6/2
x=3
Compare your solution to the restricted value.
Since the solution is the same as the restricted value we must exclude it as a solution.  Since all of the solutions have been excluded, there is no solution to the rational equation.

Video Example:

1.1 Linear Equations, Solving Equations

Higher Order Equation that reduces to a linear equation

Example: Solve the equation.

(x+5)^3-9=x(x+7)(x+8)-6

Solution:

The original equation
(x+5)^3-9=x(x+7)(x+8)-6
Simplify both sides of the equation.  On the left hand side, rewrite the exponent.  On the right hand side, begin to simplify the multiplication.
(x+5)(x+5)(x+5)-9=x(x^2+8x+7x+56)-6
Simplify both sides of the equation.  On the left hand side, begin multiplying.  On the right hand side, combine like terms.
(x+5)(x^2+5x+5x+25)-9=x(x^2+15x+56)-6
Simplify both sides of the equation.  On the left hand side, combine like terms.  On the right hand side use the distributive property.
(x+5)(x^2+10x+25)-9=x^3+15x^2+56x-6
Simplify both sides of the equation.  On the left hand side, continue multiplying.  The right hand side is in simplest form.
x(x^2+10x+25)+5(x^2+10x+25)-9=x^3+15x^2+56x-6
x^3+10x^2+25x+5x^2+50x+125-9=x^3+15x^2+56x-6
Simplify both sides of the equation.  On the left hand side, combine like terms.  The right hand side is in simplest form.
x^3+15x^2+75x+116=x^3+15x^2+56x-6
Now that each side is in simplest form we want the terms with x on one side and the constant terms on the the other side.  Subtract x^3 from each side.  It cancels from each side.
x^3-x^3+15x^2+75x+116=x^3-x^3+15x^2+56x-6
15x^2+75x+116=15x^2+56x-6
Subtract 15x^2 from each side.  It cancels from each side.
15x^2-15x^2+75x+116=15x^2-15x^2+56x-6
75x+116=56x-6
Subtract 56x from each side and simplify. 
75x-56x+116=56x-56x-6
19x+116=-6
Subtract 116 from each side and simplify.
19x+116-116=-6-116
19x=-122
Get x by it self by dividing by 19 on both sides and simplify.
{19x}/19={-122}/19
x={-122}/19
1.6 Other Types of Equations, Equations, Equations that are Quadratic in Form

Quadratic in Form (U-substitution)

Example: Solve the equation.

5x^{2/3}-6x^{1/3}+1=0

Solution:

The equation is similar to a quadratic.  It has 3 terms and one exponent is twice the other.  Since the equation is quadratic in form, use substitution to solve the equation.

Use the following substitution to rewrite the equation

u=x^{1/3}

u^2=x^{2/3}

Original Equation
5x^{2/3}-6x^{1/3}+1=0
Substitute
u=x^{1/3}
u^2=x^{2/3}
5x^{2/3}-6x^{1/3}+1=0
5u^{2}-6u+1=0
Solve the quadratic equation by factoring.
1) Factor the quadratic
5u^{2}-6u+1=0
(5u-1)(u-1)=0
Solve the quadratic equation by factoring.
2) Apply the zero product property
(5u-1)(u-1)=0
5u-1=0 or u-1=0
Solve the quadratic equation by factoring.
3) Solve each linear factor
5u-1=0 or u-1=0
5u-1+1=0+1 or u-1+1=0+1
5u=1 or u=1
{5u}/5=1/5 or u=1
u=1/5 or u=1
Substitute again to bring back the original variable.  Use the original substitution.
u=x^{1/3}
u=1/5 or u=1
x^{1/3}=1/5 or x^{1/3}=1 
Solve the equation with rational exponents.
1) Rewrite the rational exponents in radical form
x^{1/3}=1/5 or x^{1/3}=1
root{3}{x}=1/5 or root{3}{x}=1
Solve the equation with rational exponents.
2) Cancel the cube root by cubing both sides.
3) Simplify
root{3}{x}=1/5 or root{3}{x}=1
(root{3}{x})^3=(1/5)^3 or (root{3}{x})^3=(1)^3
x=1/125 or x=1

The solution to 5x^{2/3}-6x^{1/3}+1=0 isx=1/125 or x=1.

 

Here is a video with similar examples.