Mastering Log Inequalities: Solving Logarithmic Equations and Inequalities

	Example Equation
With One Variable	$2^{x} = 32$
With the Same Variable on Both Sides	$2^{2 x} = 5 \cdot 2^{x}$
With the Same Base	$4^{3 x} = 4^{2 x + 3}$
With Unlike Bases	$3^{x + 4} = 8 1^{x}$
With a Rational Base	$(\frac{1}{2})^{x} = 8$

Example Equation

With One Variable

2^{x} = 32

With the Same Variable on Both Sides

2^{2 x} = 5 \cdot 2^{x}

With the Same Base

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

With Unlike Bases

3^{x + 4} = 8 1^{x}

With a Rational Base

(\frac{1}{2})^{x} = 8

$x$	$y = ln (x - 2)$	$y = 2 lo g x$
$ln (x - 2)$	$y$	$2 lo g x$	$y$
$1$	-	-	$2 lo g 1$	$0$
$2$	-	-	$2 lo g 2$	$\approx 0.6$
$3$	$ln (3 - 2)$	$0$	$2 lo g 3$	$\approx 0.95$
$4$	$ln (4 - 2)$	$\approx 0.69$	$2 lo g 4$	$\approx 1.2$
$5$	$ln (5 - 2)$	$\approx 1.1$	$2 lo g 5$	$\approx 1.4$
$12$	$ln (12 - 2)$	$\approx 2.3$	$2 lo g 12$	$\approx 2.16$

x

y = ln (x - 2)

y = 2 lo g x

ln (x - 2)

y

2 lo g x

y

1

2 lo g 1

0

2

2 lo g 2

\approx 0.6

3

ln (3 - 2)

0

2 lo g 3

\approx 0.95

4

ln (4 - 2)

\approx 0.69

2 lo g 4

\approx 1.2

5

ln (5 - 2)

\approx 1.1

2 lo g 5

\approx 1.4

12

ln (12 - 2)

\approx 2.3

2 lo g 12

\approx 2.16

$x$	$- lo g x$	$y$
$0.1$	$- lo g 0.1$	$1$
$0.5$	$- lo g 0.5$	$\approx 0.3$
$1$	$- lo g 1$	$0$
$2$	$- lo g 2$	$\approx - 0.3$
$3$	$- lo g 3$	$\approx - 0.48$

x

- lo g x

y

0.1

- lo g 0.1

1

0.5

- lo g 0.5

\approx 0.3

1

- lo g 1

0

2

- lo g 2

\approx - 0.3

3

- lo g 3

\approx - 0.48

lo g (x - 3) < lo g (x + 6) - 1

Argument on the LHS

Argument on the RHS

x - 3 > 0

⇕

x > 3

x + 6 > 0

⇕

x > - 6

Consider the definition of a logarithm.

lo g_{b} a = c \Leftrightarrow b^{c} = a

Use this definition to solve the following equations. Round the answers to three significant digits.

2 (3)^{x} = 10

5 (\frac{1}{2})^{x + 1} = 2

Let's start by dividing both sides of the equation by 2. 2(3)^x=10 ⇔ 3^x=5 Now we can use the definition of a logarithm to rewrite the exponential equation as a logarithmic equation. Definition:& b^c= a ⇔ log_b a= c Equation:& 3^x= 5 ⇔ log_3 5= x Finally, we will use the Change of Base Formula to find the numerical value of log_3 5. Because the calculator can only calculate logarithms with base 10 or base e, we will use common logarithms, which have base 10. Let's not forget to round the answer to three significant digits!

Let's start by dividing both sides of the equation by 5. 5(1/2)^(x+1)=2 ⇔ (1/2)^(x+1)=2/5 Now we can use the definition of a logarithm to rewrite the exponential equation as a logarithmic equation. Definition b^c= a ⇔ log_b a= c [1em] Equation ( 1/2)^(x+1)= 2/5 ⇔ log_(12) 2/5= x+1 Finally, we will use the Change of Base Formula to find the numerical value of log_(12) 25. As in Part A, because the calculator can only calculate logarithms with base 10 or base e, we will use common logarithms, which have base 10. Let's not forget to round the answer to three significant digits!

Consider the Exponential Property of Inequality.

If b < 1 b^{x_{1}} \geq b^{x_{2}} ⇕ x_{1} \leq x_{2} If b > 1 b^{x_{1}} \geq b^{x_{2}} ⇕ x_{1} \geq x_{2}

Here,

b

is a positive real number different from

1 .

Use this property to solve the following inequalities.

27 (3)^{2 x} \geq \frac{3 ^{x}}{3}

\frac{1}{1 6} (4)^{x} < 6 4^{x}

Let's start by writing both sides of the inequality as a single power of 3. To do this, we will first rewrite 27 as 3^3. Then, we will use the Product of Powers Property and the Quotient of Powers Property. Let's do it!

Now, we can use the Exponential Property of Inequality. Note that the base of the powers is 3, which is greater than 1. Therefore, the expression on the left-hand side is greater than or equal to the expression on the right-hand side if and only if the exponent on the left-hand side is greater than or equal to the exponent on the right-hand side. 3^(3+2x) ≥ 3^(x-1) ⇕ 3+2x≥ x-1 We can now solve the obtained linear inequality for x.

Let's start by writing both sides of the inequality as a single power of 4. To do this, we will first rewrite 116 as 4^(- 2) and 64 as 4^3. Then, we will use the Product of Powers Property and the Power of a Power Property. Let's do it!

Now, we can use the Exponential Property of Inequality. Note that the base of the powers is 4, which is greater than 1. Therefore, similar to Part A, the expression on the left-hand side is less than the expression on the right-hand side if and only if the exponent on the left-hand side is less than the exponent on the right-hand side. 4^(- 2+x) < 4^(3x) ⇕ - 2+x< 3x Let's finally solve the linear inequality for x.

Solve the logarithmic equation graphically.

- lo g x + 3 = ln (x - 1.5)

Round the answer to the nearest integer.

To solve the logarithmic equation graphically, we will start by considering each side of the equation as a logarithmic function. - log x+3= ln (x-1.5) ⇓ y=- log x+3 and y=ln (x-1.5) Recall that the argument of a logarithm must be positive. With this information, we can write the domain of each function.

Function	Positive argument	Domain
y=- log x+3	x>0	x>0
y=ln (x-1.5)	x-1.5>0	x>1.5

Keeping in mind the domain of each function, we can make a table of values to find points on each curve. To do this, we need to use a calculator to calculate common and natural logarithms.

	y=- log x+3		y=ln (x-1.5)
x	- log x+3	y	ln (x-1.5)	y
0.5	- log 0.5+3	≈ 3.3	-	-
1	- log 1+3	3	-	-
2	- log 2+3	≈ 2.7	ln( 2-1.5)	≈ - 0.7
3	- log 3+3	≈ 2.5	ln( 3-1.5)	≈ 0.4
4	- log 4+3	≈ 2.4	ln( 4-1.5)	≈ 0.9
5	- log 5+3	≈ 2.3	ln( 5-1.5)	≈ 1.3
10	- log 10+3	2	ln( 10-1.5)	≈ 2.1

Let's now plot on a coordinate plane the points that we obtained in the table. We will also connect them with smooth curves.

The x-coordinate of the point of intersection is the solution to our equation.

Although we cannot determine an exact answer, we see that, rounded to the nearest integer, the solution is x≈ 9.

Consider the Property of Equality for Logarithmic Equations.

lo g_{b} x = lo g_{b} y \Leftrightarrow x = y

Here,

x,

y,

and

b

are positive real numbers, with

b

different from

1 .

Use this property to solve the following logarithmic equations.

lo g (2 - x) = 2 lo g x

3 ln 3 x = ln (2 x - 4)

The left-hand side of the logarithmic equation is already expressed as a single logarithm. To express the right-hand side as a single logarithm, we can use the Power Property of Logarithms. Then, we will use the Property of Equality for Logarithmic Equations to find the value of x. Let's do it!

We now have a quadratic equation. Let's identify the values of a, b, and c. x^2+x-2=0 ⇕ 1x^2+ 1x+( - 2)=0 We can find the value of x by substituting these values into the Quadratic Formula.

Finally, we will see if any of the obtained solutions are an extraneous solution. To do so, we will substitute x_1=1 and x_2=- 2 in the given equation. Let's start with x_1=1.

We obtained a true statement. This means that x_1=1 is a solution to the equation. Let's check x_2=- 2 by following the same procedure.

Note that in the right-hand side of the equation we have a logarithm with a negative argument. Since the argument of a logarithm must always be non-negative, x_2=- 2 is an extraneous solution. Therefore, the given equation has only one solution which is x=1.

This time, the right-hand side of the equation is expressed as a single logarithm. However, we need to rewrite the left-hand side as a single logarithm. To do this, we will use the Power Property of Logarithms. Keep in mind that (sqrt(x))^3 is equal to x.

Now that both sides are expressed as single logarithms, we can use the Property of Equality for Logarithmic Equations, and solve for x.

Finally, let's see if x=4 is an extraneous solution.

We obtained a true statement. Therefore, x=4 is a solution to the given equation.

Solve the logarithmic equation.

3 lo g_{5} (2 x + 1) = 15

To use the definition of a logarithm, we need to isolate the logarithmic expression on one side. To do this, we will divide both sides of the equation by 3. 3log_5 (2x+1)=15 ⇕ log_5 (2x+1)=5 Now, to continue solving the logarithmic equation, we can use the definition of a logarithm. Definition:& log_b a= c ⇔ b^c= a Equation:& log_5 ( 2x+1)= 5 ⇔ 5^5= 2x+1 We obtained a linear equation. Let's solve it for x.

Let's check if x=1562 is an extraneous solution.

Since we obtained a true statement, x=1562 is a solution to the equation.

Solve the logarithmic inequality.

lo g_{2} (x - 4) < lo g_{2} (2 x + 8) - 2

Write the answer in interval notation.

We want to solve an inequality. To do so, we will start by writing 2 as a logarithm with base 2. Let's use the definition of a logarithm. 2^2= 4 ⇔ log_2 4= 2 Therefore, in the given inequality, we can substitute log_2 4 for 2. log_2 (x-4) < log_2(2x+8) - 2 ⇕ log_2 (x-4) < log_2(2x+8) - log_2 4 Next, we can use the Quotient Property of Logarithms to express the right-hand side as a single logarithm. log_2 (x-4) < log_2(2x+8) - log_2 4 ⇕ log_2 (x-4) < log_2 2x+8/4 Now both sides are expressed as a single logarithm with the same base. Therefore, by the Logarithmic Property of Inequality, we can eliminate the logarithms. Note that the base of the logarithms is 2, which is greater than 1. log_2 (x-4) < log_2 2x+8/4 ⇕ x-4< 2x+8/4 Let's solve the obtained linear inequality.

Finally, recall that logarithms are only defined for positive numbers. Therefore, the arguments of the logarithms in the original inequality must be greater than zero.

log_2 (x-4) < log_2(2x+8) - 2
Argument on the LHS	Argument on the RHS
x-4>0 ⇕ x>4	2x+8>0 ⇕ x>- 4

The solution set is the intersection of three inequalities. Inequality I:& x<12 Inequality II:& 4< x Inequality III:& - 4number line.

The solution to the inequality are all real numbers greater than 4 and less than 12. This solution can be written as a compound inequality or in interval notation. 4

Solve the exponential equation.

4 (2)^{2 x} = \frac{3 ^{x}}{9}

Round the answer to one decimal place.

To solve the exponential equation, we will start by writing both sides as single powers. To do so, we will rewrite 4 as 2^2 and 9 as 3^2, and use the Product of Powers Property and the Quotient of Powers Property.

Now that both sides are expressed as single powers, we can take logarithms on both sides. Then, to remove the variable from the exponents, we will use the the Power Property of Logarithms.

Finally, we can solve the obtained linear equation for x. Let's. do it!

	15 Theory slides
	13 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

$lo g (x - 3) < lo g (x + 6) - 1$
Argument on the LHS	Argument on the RHS
$x - 3 > 0$ $⇕$ $x > 3$	$x + 6 > 0$ $⇕$ $x > - 6$

Solving Logarithmic Equations and Inequalities

Catch-Up and Review

Alternative Solution

Hint

Solution

Proof

b>1

b<1

Hint

Solution

Hint

Solution

Proof

logb​x=logb​y⇒ x=y

x=y⇒ logb​x=logb​y

Hint

Solution

Hint

Solution

Proof

b>1

b<1

Hint

Solution

Hint

Solution

Solving Logarithmic Equations and Inequalities

Recommended exercises

$b > 1$

$b < 1$

$lo g_{b} x = lo g_{b} y \Rightarrow x = y$

$x = y \Rightarrow lo g_{b} x = lo g_{b} y$

$b > 1$

$b < 1$