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§ 9.5 Exponential and Logarithmic Equations

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Blitzer, Intermediate Algebra, 5e – Slide #2 Section 9.5 Exponential Equations At age 20, you inherit $30,000. You’d like to put aside $25,000 and eventually have over half a million dollars for early retirement. Is this possible? In this section, you will see how techniques for solving equations with variable exponents provide an answer to this question. One technique involves using logarithms to solve the exponential equations. Remember that you are solving for x and that the x in these equations is in the exponent. So this is different than any equation we have looked at before. And for the answer to the question on when the $25,000 would grow to $500,000 – If the money was invested at 9% compounded monthly, it would grow to the half million dollars you wanted for your retirement in 33.4 years.

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Blitzer, Intermediate Algebra, 5e – Slide #3 Section 9.5 Exponential Equations 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.

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Blitzer, Intermediate Algebra, 5e – Slide #4 Section 9.5 Exponential EquationsEXAMPLE Solve: SOLUTION This is the given equation. In each equation, express both sides as a power of the same base. Then set the exponents equal to each other. (a) Because 27 is, we express each side of the equation in terms of the base, 3. Write each side as a power of the same base. Equate the exponents. Subtract 1 from both sides.

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Blitzer, Intermediate Algebra, 5e – Slide #5 Section 9.5 Exponential Equations Divide both sides by 2. CONTINUED Check 1: ? ? This is the given equation. (b) Because can both be expressed as 7 raised to a power, we will write each side of the equation in terms of 7. The solution is 1 and the solution set is {1}. true

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Blitzer, Intermediate Algebra, 5e – Slide #6 Section 9.5 Exponential EquationsCONTINUED Write each side as a power of the same base. Equate the bases. Multiply both sides by the LCD, 6. Add 2 to both sides. Check 5: ?

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Blitzer, Intermediate Algebra, 5e – Slide #7 Section 9.5 Exponential EquationsCONTINUED ? ? The solution is 5 and the solution set is {5}. true

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Blitzer, Intermediate Algebra, 5e – Slide #8 Section 9.5 Exponential Equations Using Natural Logarithms to Solve Exponential Equations 1) Isolate the exponential expression. 2) Take the natural logarithm on both sides of the equation. 3) Simplify using one of the following properties: 4) Solve for the variable.

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Blitzer, Intermediate Algebra, 5e – Slide #9 Section 9.5 Exponential EquationsEXAMPLE Solve: SOLUTION This is the given equation. We begin by dividing both sides by 1250 to isolate the exponential expression,. Then we take the natural logarithm on both sides of the equation. Isolate the exponential factor by dividing both sides by Take the natural logarithm on both sides.

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Blitzer, Intermediate Algebra, 5e – Slide #10 Section 9.5 Exponential Equations Use the inverse property CONTINUED on the left side. Divide both sides by Thus, the solution of the equation is. Try checking this approximate solution in the original equation, verifying that is the solution set.

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Blitzer, Intermediate Algebra, 5e – Slide #11 Section 9.5 Logarithmic EquationsEXAMPLE Solve: SOLUTION This is the equivalent equation. We first rewrite the equation as an equivalent equation in exponential form using the fact that Now we solve the equivalent equation for x. Evaluate the exponent. Subtract 1 from both sides. Divide both sides by 4.

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Blitzer, Intermediate Algebra, 5e – Slide #12 Section 9.5 Logarithmic Equations This is the given logarithmic equation. CONTINUED Check 31/4: Replace x with 31/4. Multiply. Add. ? ? ? true This true statement indicates that the solution is 31/4 and the solution set is {31/4}.

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Blitzer, Intermediate Algebra, 5e – Slide #13 Section 9.5 Logarithmic EquationsEXAMPLE Solve: SOLUTION This is the given equation. Use the product rule to obtain a single logarithm. Multiply. Evaluate the exponent. Subtract 8 from both sides. Factor.

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Blitzer, Intermediate Algebra, 5e – Slide #14 Section 9.5 Logarithmic Equations Set each factor equal to 0. CONTINUED Solve for x. Check 3:Check -3: Negative numbers do not have logarithms. false ?? ? ? ? ?

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Blitzer, Intermediate Algebra, 5e – Slide #15 Section 9.5 Logarithmic EquationsCONTINUED true The solutions are 3 and -3. The solution set is {3, -3}. This is the given equation. Use the quotient rule to obtain a single logarithm. Evaluate the exponent. Multiply both sides by the LCD, x.

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Blitzer, Intermediate Algebra, 5e – Slide #16 Section 9.5 Logarithmic EquationsCONTINUED Subtract 2x from both sides. Divide both sides by 98. Check -1/98: Negative numbers do not have logarithms. Therefore there is no solution and the solution set is. false

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Blitzer, Intermediate Algebra, 5e – Slide #17 Section 9.5 Logarithmic EquationsEXAMPLE Solve: SOLUTION This is the given equation. Subtract 7 from both sides. Divide both sides by 3. Exponentiate both sides. Simplify the left side. Evaluate the exponent.

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Blitzer, Intermediate Algebra, 5e – Slide #18 Section 9.5 Logarithmic Equations Check : CONTINUED This is the given equation. Replace x with Because, we conclude Multiply. Subtract. This true statement indicates that the solution is and the solution set is ? ? ? true

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Blitzer, Intermediate Algebra, 5e – Slide #19 Section 9.5 Exponential Equations in ApplicationEXAMPLE The formula models the population of Florida, A, in millions, t years after When will the population of Florida reach 19.2 million? SOLUTION This is the given equation. We will replace A with 19.2 and then solve for t. Replace A with Divide both sides by Take the natural logarithm of both sides.

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Blitzer, Intermediate Algebra, 5e – Slide #20 Section 9.5 Exponential Equations in Application Use the inverse property on the right side. CONTINUED Divide both sides by The state of Florida will have a population of 19.2 million people approximately 7.76 years after the year 2000.

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