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Solving Exponential Equations

Solving Exponential Equations. Equations with variables in exponents, such as 3 x = 5 and 7 3 x = 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. Example.

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Solving Exponential Equations

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  1. Solving Exponential Equations Equations with variables in exponents, such as 3x = 5 and 73x = 90 are called exponential equations. In Section 9.3, we solved certain logarithmic equations by using the principle am = x means logax = m

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

  3. Solve graphically

  4. Example Solve: e1.32t = 2000 Solution e1.32t = 2000 We have: Note that we use the natural logarithm lne1.32t = ln 2000 Logarithmic and exponential functions are inverses of each other 1.32t = ln 2000

  5. To Solve an Equation of the Form at = 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.

  6. Example Solve: log2(6x + 5) = 4. Solving Logarithmic Equations Solution log2(6x + 5) = 4 logax = m means am = x 6x + 5 = 24 6x + 5 = 16 6x = 11 x = 11/6 The solution is x = 11/6.

  7. Using change of base Solve graphically x = 11/6

  8. Example Solve: log x + log(x + 9) = 1. Solution To increase the understanding, we write in the base 10. log10x + log10 (x + 9) = 1 x = 1: log10[x(x + 9)] = 1 log 1 + log(1 + 9) = 1 x(x + 9) = 101 0 + log(10) = 1 0 + 1 = 1 TRUE x2+ 9x = 10 The logarithm of a negative x = –10: x2+ 9x – 10 = 0 number is undefined. FALSE log (–10) + log(–10 + 9) = 1 (x – 1)(x + 10) = 0 x – 1 = 0 or x + 10 = 0 x = 1 or x = –10 The solution is x = 1.

  9. Solve graphically We graph y1= log(x) + log(x + 9) - (1) x = 1

  10. Example Solve: log3(2x + 3) – log3(x – 1) = 2. Solution log3(2x + 3) – log3(x – 1) = 2 (2x + 3) = 9(x – 1) 2x + 3 = 9x – 9 x = 12/7 The solution is 12/7. Check is left to the student.

  11. Solution

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