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Solving Exponential Equations Equations with variables in exponents, such as 3 x = 5 and 7 3x = 90 are called exponential equations. In Section 9.3, we.

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Presentation on theme: "Solving Exponential Equations Equations with variables in exponents, such as 3 x = 5 and 7 3x = 90 are called exponential equations. In Section 9.3, we."— Presentation transcript:

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2 Solving Exponential Equations Equations with variables in exponents, such as 3 x = 5 and 7 3x = 90 are called exponential equations. In Section 9.3, we solved certain logarithmic equations by using the principle a m = x means log a x = m

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5 Solution Example Solve: 3 x +1 = 43 We have 3 x +1 = 43 log 3 x +1 = log 43 (x +1)log 3 = log 43 Principle of logarithmic equality Power rule for logs Shuhaw Answer Chemistry Answer

6 Solve graphically

7 Solution Example Solve: e 1.32t = 2000 We have: Note that we use the natural logarithm Logarithmic and exponential functions are inverses of each other e 1.32t = 2000 ln e 1.32t = ln t = ln 2000

8 To Solve an Equation of the Form a t = b for t 1.Take the logarithm (either natural or common) of both sides. 2.Use the power rule for exponents so that the variable is no longer written as an exponent. 3.Divide both sides by the coefficient of the variable to isolate the variable. 4.If appropriate, use a calculator to find an approximate solution in decimal form.

9 Solution Example Solve: log 2 (6x + 5) = 4. 6x + 5 = 2 4 6x = 11 The solution is x = 11/6. log 2 (6x + 5) = 4 6x + 5 = 16 x = 11/6 log a x = m means a m = x Solving Logarithmic Equations

10 Solve graphically x = 11/6 Using change of base

11 Solution Example Solve: log x + log (x + 9) = 1. x 2 + 9x = 10 To increase the understanding, we write in the base 10. log 10 x + log 10 (x + 9) = 1 log 10 [x(x + 9)] = 1 x(x + 9) = 10 1 x 2 + 9x – 10 = 0 (x – 1)(x + 10) = 0 x – 1 = 0 or x + 10 = 0 x = 1 or x = – log (10) = 1 x = 1: log 1 + log (1 + 9) = = 1 TRUE x = –10: log (–10) + log (–10 + 9) = 1 FALSE The solution is x = 1. The logarithm of a negative number is undefined.

12 Solve graphically We graph y 1 = log(x) + log (x + 9)  (1) x = 1

13 Solution Example Solve: log 3 (2x + 3) – log 3 (x – 1) = 2. log 3 (2x + 3) – log 3 (x – 1) = 2 (2x + 3) = 9(x – 1) x = 12/7 2x + 3 = 9x – 9 The solution is 12/7. Check is left to the student.

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