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Exponential and Logarithmic Equations

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Presentation on theme: "Exponential and Logarithmic Equations"β€” Presentation transcript:

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

2 Vocabulary exponential equation logarithmic equation

3 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

4 Rewrite the equation in the form 𝑏 π‘₯ = 𝑏 𝑦 . Set π‘₯=𝑦.
Solving Exponential Equations by Expressing Each Side as a Power of the Same Base Express each side as a power of the same base. Set the exponents equal to each other. Rewrite the equation in the form 𝑏 π‘₯ = 𝑏 𝑦 . Set π‘₯=𝑦. Solve for the variable.

5 Solving Exponential Equations
Solve and check. 98 – x = 27x – 3 Rewrite each side with the same base; 9 and 27 are powers of 3. (32)8 – x = (33)x – 3 To raise a power to a power, multiply exponents. 316 – 2x = 33x – 9 Bases are the same, so the exponents must be equal. 16 – 2x = 3x – 9 x = 5 Solve for x.

6 Check 98 – x = 27x – 3 98 – – 3 οƒΌ The solution is x = 5.

7 Solve and check: 𝟐 πŸ‘π’™βˆ’πŸ– =πŸπŸ” 2 3π‘₯βˆ’8 = 2 4 3π‘₯βˆ’8=4 π‘₯=4
Rewrite each side with the same base. 2 3π‘₯βˆ’8 = 2 4 Now that the bases are the same, solve for π‘₯ 3π‘₯βˆ’8=4 π‘₯=4

8 Check: 𝟐 πŸ‘π’™βˆ’πŸ– =πŸπŸ” Substitute 4 for the variable. 2 3(4)βˆ’8 =16 2 12βˆ’8 =16 2 4 =16 16=16

9 Solve and check: πŸ“ πŸ‘π’™βˆ’πŸ” =πŸπŸπŸ“ 5 3π‘₯βˆ’6 = 5 3 3π‘₯βˆ’6=3 π‘₯=3
Rewrite each side with the same base. 5 3π‘₯βˆ’6 = 5 3 Now that the bases are the same, solve for π‘₯ 3π‘₯βˆ’6=3 π‘₯=3

10 Solve and check: πŸ“ πŸ‘π’™βˆ’πŸ” =πŸπŸπŸ“ 5 3(3)βˆ’6 =125 5 9βˆ’6 =125 5 3 =125 125=125
Substitute 3 for the variable and solve. 5 3(3)βˆ’6 =125 5 9βˆ’6 =125 5 3 =125 125=125

11 Solving Exponential Equations
Solve and check. 4x – 1 = 5 log 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. x –1 = log5 log4 Divide both sides by log 4. x = β‰ˆ 2.161 log5 log4 Check Use a calculator. The solution is x β‰ˆ

12 Using Logarithms to Solve Exponential Equations
Isolate the exponential expression. 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. Simplify using one of the following properties: π₯𝐧 𝒃 𝒙 =𝒙 π₯𝐧 𝒃 or π₯𝐧 𝒆 𝒙 =𝒙 or π₯𝐨𝐠 𝟏𝟎 𝒙 =𝒙 Solve for the equation.

13 Solve: πŸ’ 𝒙 =πŸπŸ“ ln 4 π‘₯ = ln 15 Take the natural logarithm on both sides π‘₯ ln 4= ln 15 Use the power rule π‘₯ = ln 15 ln 4 Solve for x by dividing both sides by π₯𝐧 πŸ’ π’™β‰ˆπŸ.πŸ—πŸ“πŸ‘πŸ’ Use calculator. β‰ˆ14.999β‰ˆ15 Check.

14 Solve: 𝑒 π‘₯ =72 ln 𝑒 π‘₯ = ln 72 π‘₯ ln 𝑒= ln 72 π‘₯= ln 72 π‘₯β‰ˆ4.277
Take the natural logarithm of both sides. ln 𝑒 π‘₯ = ln 72 π‘₯ ln 𝑒= ln 72 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 𝒍𝒏 𝒆 𝒙 =𝒙 . Also, ln 𝑒=1 π‘₯= ln 72 π‘₯β‰ˆ4.277 Check your answer.

15 Solve: πŸ’πŸŽ 𝒆 𝟎.πŸ”π’™ βˆ’πŸ‘=πŸπŸ‘πŸ• 40 𝑒 0.6π‘₯ =240 Add 3 to both sides 𝑒 0.6π‘₯ =6 Divide both sides by 40 ln 𝑒 0.6π‘₯ = ln 6 Take the natural logarithm of both sides 0.6π‘₯= ln 6 Use the inverse property 𝒍𝒏 𝒆 𝒙 =𝒙 . 𝒙= π₯𝐧 πŸ” 𝟎.πŸ” β‰ˆ 2.99 Divide both sides by 0.6 and solve for x

16 Solve πŸ“ π’™βˆ’πŸ = πŸ’ πŸπ’™+πŸ‘ ln 5 π‘₯βˆ’2 = ln 4 2π‘₯+3 Take the natural logarithm on both sides π‘₯βˆ’2 ln 5 =(2π‘₯+3) ln 4 Use the power rule π‘₯ ln 5 βˆ’2 ln 5 =2π‘₯ ln 4 +3 ln 4 Use the distributive property π‘₯ ln 5 βˆ’2π‘₯ ln 4 =2 ln 5 +3 ln 4 Rearrange terms π‘₯( ln 5βˆ’2 ln 4)= 2 ln 5 +3 ln 4 Factor out x 2 ln 5+3 ln 4 π‘₯= ln 5βˆ’2 ln 4 The solution is approximately βˆ’πŸ”.πŸ‘πŸ’

17 Solve: 𝒆 πŸπ’™ βˆ’πŸ’ 𝒆 𝒙 +πŸ‘=𝟎 Let 𝒖= 𝒆 𝒙 𝑒 2 βˆ’4𝑒+3=0 Substitute 𝒖 for 𝒆 𝒙
π‘’βˆ’3 π‘’βˆ’1 =0 Factor on the left π‘’βˆ’3=0 π‘œπ‘Ÿ π‘’βˆ’1=0 Set each factor equal to 0 𝑒= 𝑒=1 Solve for 𝒖 𝑒 π‘₯ = 𝑒 π‘₯ =1 Replace 𝒆 𝒙 for 𝒖 Take the natural logarithm of both sides ln 𝑒 π‘₯ = ln ln 𝑒 π‘₯ = ln 1 𝒙= π₯𝐧 πŸ‘ β‰ˆπŸ.𝟏𝟎 𝒙=𝟎

18 Using the Definition of a Logarithm to Solve Logarithmic Equations
Express the equation in the form log 𝑏 𝑀=𝑐. Use the definition of a logarithm to rewrite the equation in exponential form. π₯𝐨𝐠 𝐛 𝐌=𝐜 means 𝐛 𝐜 =𝐌 Solve for the variable. Check your solutions for 𝑀>0 in the original equation.

19 Solve and check: π₯𝐨𝐠 πŸ’ 𝒙+πŸ‘ =𝟐
4 2 =π‘₯+3 Rewrite in exponential form 16=π‘₯+3 πŸπŸ‘=𝒙

20 Solve and check: πŸ‘ π₯𝐧 πŸπ’™ =𝟏𝟐 ln 2π‘₯ =4 Divide both sides by 3 Rewrite the natural logarithm showing base e log 𝑒 2π‘₯ =4 𝑒 4 =2π‘₯ Rewrite in exponential form 𝒆 πŸ’ 𝟐 =π’™β‰ˆπŸπŸ•.πŸ‘ Divide both sides by 2

21 Solve and check: π₯𝐨𝐠 𝟐 𝒙 + π₯𝐨𝐠 𝟐 π’™βˆ’πŸ• =πŸ‘
log 2 [π‘₯ π‘₯βˆ’7 ]=3 Use the product rule 2 3 =π‘₯(π‘₯βˆ’7) Rewrite in exponential form 8= π‘₯ 2 βˆ’7π‘₯ Simplify π‘₯ 2 βˆ’7π‘₯βˆ’8=0 Set up as a quadratic equation π‘₯βˆ’8 π‘₯+1 =0 Factor π‘₯βˆ’8=0  𝒙=πŸ– π‘₯+1=  𝒙=βˆ’πŸ Always check your answers with original equation.

22 If π₯𝐨𝐠 𝒃 𝑴= π₯𝐨𝐠 𝒃 𝑡 , then 𝑴=𝑡.
Using the One-to-One Property of Logarithms to Solve Logarithmic Equations Express the equation in the form log 𝑏 𝑀= log 𝑏 𝑁. The coefficient must be equal to 1 on both sides. Use the one-to-one property to rewrite the equation without the logarithm. If π₯𝐨𝐠 𝒃 𝑴= π₯𝐨𝐠 𝒃 𝑡 , then 𝑴=𝑡. Solve for the variable. Check proposed solutions in the original equation. 𝑀 and 𝑁 must be positive.

23 ( π‘₯+2 4π‘₯+3 ) = ln ( 1 π‘₯ ) ln Solve: π₯𝐧 𝒙+𝟐 βˆ’ π₯𝐧 πŸ’π’™+πŸ‘ = π₯𝐧 ( 𝟏 𝒙 )
Use the quotient rule π‘₯+2 4π‘₯+3 = 1 π‘₯ Use the one-to-one property π‘₯ π‘₯+2 =1(4π‘₯+3) Cross multiply π‘₯ 2 +2π‘₯=4π‘₯+3 Use distributive property π‘₯ 2 +2π‘₯βˆ’4π‘₯βˆ’3=0 Set up as a quadratic equation π‘₯ 2 βˆ’2π‘₯βˆ’3=0

24 Check by substituting each solution into the original equation.
π‘₯ 2 βˆ’2π‘₯βˆ’3=0 (π‘₯βˆ’3)(π‘₯+1) Factor π‘₯βˆ’3= π‘₯+1=0 Set each factor equal to 0 𝒙=πŸ‘ 𝒙=βˆ’πŸ Check by substituting each solution into the original equation. π₯𝐧 𝒙+𝟐 βˆ’ π₯𝐧 πŸ’π’™+πŸ‘ = π₯𝐧 ( 𝟏 𝒙 )

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