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§ 9.5 Exponential and Logarithmic Equations

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**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. Blitzer, Intermediate Algebra, 5e – Slide #2 Section 9.5

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

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**Exponential Equations**

EXAMPLE Solve: SOLUTION 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. This is the given equation. Write each side as a power of the same base. Equate the exponents. Subtract 1 from both sides. Blitzer, Intermediate Algebra, 5e – Slide #4 Section 9.5

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**Exponential Equations**

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

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**Exponential Equations**

CONTINUED 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: ? Blitzer, Intermediate Algebra, 5e – Slide #6 Section 9.5

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**Exponential Equations**

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

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**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. Blitzer, Intermediate Algebra, 5e – Slide #8 Section 9.5

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**Exponential Equations**

EXAMPLE Solve: SOLUTION 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. This is the given equation. Isolate the exponential factor by dividing both sides by 1250. Take the natural logarithm on both sides. Blitzer, Intermediate Algebra, 5e – Slide #9 Section 9.5

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**Exponential Equations**

CONTINUED Use the inverse property 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. Blitzer, Intermediate Algebra, 5e – Slide #10 Section 9.5

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**Logarithmic Equations**

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

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**Logarithmic Equations**

CONTINUED Check 31/4: This is the given logarithmic equation. ? 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}. Blitzer, Intermediate Algebra, 5e – Slide #12 Section 9.5

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**Logarithmic Equations**

EXAMPLE 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. Blitzer, Intermediate Algebra, 5e – Slide #13 Section 9.5

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**Logarithmic Equations**

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

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**Logarithmic Equations**

CONTINUED 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. Blitzer, Intermediate Algebra, 5e – Slide #15 Section 9.5

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**Logarithmic Equations**

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

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**Logarithmic Equations**

EXAMPLE 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. Blitzer, Intermediate Algebra, 5e – Slide #17 Section 9.5

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**Logarithmic Equations**

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

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**Exponential Equations in Application**

EXAMPLE The formula models the population of Florida, A, in millions, t years after When will the population of Florida reach 19.2 million? SOLUTION We will replace A with 19.2 and then solve for t. This is the given equation. Replace A with 19.2. Divide both sides by 15.9. Take the natural logarithm of both sides. Blitzer, Intermediate Algebra, 5e – Slide #19 Section 9.5

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**Exponential Equations in Application**

CONTINUED Use the inverse property on the right side. 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. Blitzer, Intermediate Algebra, 5e – Slide #20 Section 9.5

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In this section we will introduce a new concept which is the logarithm

In this section we will introduce a new concept which is the logarithm

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