objectives solve exponential and logarithmic equations and equalities. solve problems involving...

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Exponential and Logarithmic Equations Objectives Solve exponential and logarithmic equations and equalities. Solve problems involving exponential and logarithmic equations.

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Exponential and Logarithmic Equations

Objectives

Solve exponential and logarithmic equations and equalities.

Solve problems involving exponential and logarithmic equations.

exponential equation

logarithmic equation

Vocabulary

An exponential equation is an equation containing one or more expressions that have a variable as an exponent. To solve exponential equations:

Try writing them so that the bases are the same

Take the logarithm of both sides

Solving Exponential Equations by Expressing Each Side as a Power of the Same Base

1. Rewrite the equation in the form 2. Set 3. Solve for the variable.

Express each side as a power of the same base.

Set the exponents equal to each other.

Solve and check.

98 – x = 27x – 3

(32)8 – x = (33)x – 3 Rewrite each side with the same base; 9 and 27 are powers of 3.

316 – 2x = 33x – 9 To raise a power to a power, multiply exponents.

Solving Exponential Equations

16 – 2x = 3x – 9Bases are the same, so the exponents must be equal.

x = 5 Solve for x.

Check

98 – x = 27x – 3

98 – 5 275 – 3

93 272

729 729

The solution is x = 5.

Solve and check:

Rewrite each side with the same base.2

3 π‘₯βˆ’ 8=24

Now that the bases are the same, solve for

3 π‘₯βˆ’8=4π‘₯=4

Check:

Substitute 4 for the variable.2

3 (4 )βˆ’8=16

212βˆ’8=16

24=16

16=16

Solve and check:

Rewrite each side with the same base. 53π‘₯βˆ’6=53

Now that the bases are the same, solve for 3 π‘₯βˆ’6=3

π‘₯=3

Solve and check:

Substitute 3 for the variable and solve. 53(3)βˆ’ 6=125

59βˆ’6=125

53=125

125=125

Solve and check.

4x – 1 = 5log 4x – 1 = log 5 We cannot get common bases,

so take the log of both sides.(x – 1)log 4 = log 5 Apply the Power Property of

Logarithms.

Solving Exponential Equations

Divide both sides by log 4.

Check Use a calculator.

The solution is x β‰ˆ 2.161.

x = 1 + β‰ˆ 2.161log5log4

x –1 = log5log4

Using Logarithms to Solve Exponential Equations

1. Isolate the exponential expression.2. Take the natural logarithm on both sides of the

equation for bases other than 10. Take the common logarithm on both sides of the equation for base 10.

3. Simplify using one of the following properties:

or or

4. Solve for the equation.

Solve:

ln 4π‘₯= ln 15 Take the natural logarithm on both sides

π‘₯ ln 4=ln15 Use the power rule

= Solve for x by dividing both sides by

π’™β‰ˆπŸ .πŸ—πŸ“πŸ‘πŸ’ Use calculator.

Check.41.9534 β‰ˆ14.999β‰ˆ15

When you take the natural logarithm of base e, the ln e drops from the equation leaving only the exponent as seen above. This is using the inverse property Also,

Solve:

Take the natural logarithm of both sides.ln𝑒π‘₯=ln 72

π‘₯ ln𝑒=ln 72

π‘₯=ln 72

π‘₯β‰ˆ 4.277

Check your answer.

Solve:

40𝑒0.6π‘₯=240 Add 3 to both sides

𝑒0.6 π‘₯=6 Divide both sides by 40

ln𝑒0.6 π‘₯=ln 6 Take the natural logarithm of both sides

0.6 π‘₯=ln 6 Use the inverse property

2.99 Divide both sides by 0.6 and solve for x

Solve

ln 5π‘₯βˆ’2=ln 42π‘₯+3 Take the natural logarithm on both sides

(π‘₯βˆ’2 ) ln5=(2 π‘₯+3) ln 4 Use the power rule

π‘₯ ln5βˆ’2 ln 5=2π‘₯ ln 4+3 ln 4 Use the distributive property

π‘₯ ln5βˆ’2π‘₯ ln 4=2 ln5+3 ln 4 Rearrange terms

Factor out x

π‘₯=ΒΏ2 ln 5+3 ln 4ln 5βˆ’2 ln 4

The solution is approximately

Solve: Let

(π‘’βˆ’3 ) (π‘’βˆ’1 )=0 Factor on the left

π‘’βˆ’3=0π‘œπ‘Ÿπ‘’βˆ’1=0 Set each factor equal to 0

𝑒=3𝑒=1 Solve for

Take the natural logarithmof both sides

𝑒2βˆ’4𝑒+3=0 Substitute for

Replace for

Using the Definition of a Logarithmto Solve Logarithmic Equations

1. Express the equation in the form 2. Use the definition of a logarithm to rewrite the

equation in exponential form.

means

3. Solve for the variable.4. Check your solutions for in the original equation.

Solve and check:

42=π‘₯+3 Rewrite in exponential form

16=π‘₯+3

πŸπŸ‘=𝒙

Solve and check:

ln (2 π‘₯ )=4 Divide both sides by 3

log𝑒 (2 π‘₯ )=4 Rewrite the natural logarithm showing base e

𝑒4=2 π‘₯ Rewrite in exponential form

π’†πŸ’

𝟐=π’™β‰ˆπŸπŸ• .πŸ‘ Divide both sides by 2

Solve and check:

log 2[π‘₯ (π‘₯βˆ’7 )]=3 Use the product rule

23=π‘₯(π‘₯βˆ’7) Rewrite in exponential form

8=π‘₯2βˆ’7π‘₯ Simplify

π‘₯2βˆ’7 π‘₯βˆ’8=0 Set up as a quadratic equation

(π‘₯βˆ’8 ) (π‘₯+1 )=0 Factor

Always check your answers with original equation.

Using the One-to-One Property of Logarithms to Solve Logarithmic Equations

1) Express the equation in the form The coefficient must be equal to 1 on both sides.

2) Use the one-to-one property to rewrite the equation without the logarithm.

If , then .

3) Solve for the variable.4) Check proposed solutions in the original equation.

must be positive.

Solve:

ln() Use the quotient rule

π‘₯+24 π‘₯+3

=1π‘₯

Use the one-to-one property

π‘₯ (π‘₯+2 )=1(4 π‘₯+3) Cross multiply

π‘₯2+2π‘₯=4 π‘₯+3 Use distributive property

π‘₯2+2π‘₯βˆ’4 π‘₯βˆ’3=0

π‘₯2βˆ’2 π‘₯βˆ’3=0Set up as a quadratic equation

π‘₯2βˆ’2 π‘₯βˆ’3=0

(π‘₯βˆ’3)(π‘₯+1) Factor

𝒙=πŸ‘ 𝒙=βˆ’πŸ

Set each factor equal to 0

Check by substituting each solution into the original equation.

π₯𝐧 (𝒙+𝟐 )βˆ’π₯𝐧 (πŸ’ 𝒙+πŸ‘ )=π₯𝐧 (πŸπ’™

)