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

The equation involves natural logarithms. To solve it, the first step is to rewrite the right-hand side as a single logarithm by using the Power Property of Logarithms.

ln (x + 1) = 2 ln x

$b \cdot ln (a) = ln (a^{b})$

ln (x + 1) = ln x^{2}

Since both sides are single logarithms with the same base, the Property of Equality for Logarithmic Equations can be used.

ln (x + 1) = ln x^{2} ⇕ x + 1 = x^{2}

Notice that the resulting equation is a quadratic equation. Therefore, the values of

a,

b,

and

c

need to be identified.

x + 1 = x^{2} ⇕ - 1 x^{2} + 1 x + 1 = 0

Next,

a = - 1,

b = 1,

and

c = 1

will be substituted in the Quadratic Formula.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

x = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) ( 1 )}{2 ( - 1 )}

▼

Evaluate right-hand side

BaseOne

$1^{a} = 1$

x = \frac{- 1 \pm 1 - 4 ( - 1 ) ( 1 )}{2 ( - 1 )}

MultPosNeg

$a (- b) = - a \cdot b$

x = \frac{- 1 \pm 1 - ( - 4 ) ( 1 )}{- 2}

IdPropMult

Identity Property of Multiplication

x = \frac{- 1 \pm 1 - ( - 4 )}{- 2}

SubNeg

$a - (- b) = a + b$

x = \frac{- 1 \pm 5}{- 2}

StateSol

State solutions

x_{1} = \frac{- 1 + 5}{- 2} x_{2} = \frac{- 1 - 5}{- 2} (I) (II)

$(I), (II):$ Use a calculator

x_{1} = - 0.618033 \dots x_{2} = 1.618033 \dots

RoundSigDig

Round to $3$ significant digit(s)

x_{1} \approx - 0.618 x_{2} \approx 1.62

Finally, the solutions can be verified by substituting them into the original equation. The solution

x \approx - 0.618

will be checked first.

ln (x + 1) = 2 ln x

$x \approx - 0.618$

ln (- 0.618 + 1) \approx ? 2 ln - 0.618 \times

Logarithms are only defined for positive numbers, so

x \approx - 0.618

is an extraneous solution. The solution

x \approx 1.62

will be checked now.

ln (x + 1) = 2 ln x

$x \approx 1.62$

ln (1.62 + 1) \approx ? 2 ln 1.62

AddTerms

Add terms

ln 2.62 \approx ? 2 ln 1.62

$b \cdot ln (a) = ln (a^{b})$

ln 2.62 \approx ? ln 1.6 2^{2}

CalcPow

Calculate power

ln 2.62 \approx ? ln 2.6244

UseCalc

Use a calculator

0.963174 \dots \approx 0.964852 \dots ✓

A true statement was found this time. Therefore, the only solution to the equation is

x \approx 1.62 .

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

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

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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

$b > 1$

$b < 1$

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

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

$b > 1$

$b < 1$