This is the given equation Add 3 to both sides Isolate the exponential factor

This is the given equation add 3 to both sides

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This is the given equation. Add 3 to both sides. Isolate the exponential factor by dividing both sides by 40. Take the natural logarithm on both sides. Use the inverse property on the left. Divide both sides by 0.6 and solve for x . x = ln 6 0.6 L 2.99 ln e x = x 0.6 x = ln 6 ln e 0.6 x = ln 6 e 0.6 x = 6 40 e 0.6 x = 240 40 e 0.6 x - 3 = 237 e 0.6 x . 40 e 0.6 x - 3 = 237. EXAMPLE 3 Thus, the solution of the equation is Try checking this approximate solution in the original equation to verify that is the solution set. Check Point 3 Solve: Find the solution set and then use a calculator to obtain a decimal approximation to two decimal places for the solution. Solving an Exponential Equation Solve: Solution Because each exponential expression is isolated on one side of the equation, we begin by taking the natural logarithm on both sides. This is the given equation. Take the natural logarithm on both sides. Use the power rule and bring the variable exponents to the front: Use the distributive property to distribute ln 5 and ln 4 to both terms in parentheses. Collect variable terms involving on the left by subtracting and adding 2 ln 5 on both sides. Factor out from the two terms on the left. Isolate by dividing both sides by The solution set is The solution is approximately Check Point 4 Solve: Find the solution set and then use a calculator to obtain a decimal approximation to two decimal places for the solution. 3 2 x - 1 = 7 x + 1 . - 6.34. e 2 ln 5 + 3 ln 4 ln 5 - 2 ln 4 f . ln 5 - 2 ln 4. x x = 2 ln 5 + 3 ln 4 ln 5 - 2 ln 4 x x 1 ln 5 - 2 ln 4 2 = 2 ln 5 + 3 ln 4 2 x ln 4 x x ln 5 - 2 x ln 4 = 2 ln 5 + 3 ln 4 x ln 5 - 2 ln 5 = 2 x ln 4 + 3 ln 4 ln b x = x ln b . (x-2) ln 5=(2x+3) ln 4 Be sure to insert parentheses around the binomials. Remember that ln 5 and ln 4 are constants, not variables. ln 5 x - 2 = ln 4 2 x + 3 5 x - 2 = 4 2 x + 3 5 x - 2 = 4 2 x + 3 . EXAMPLE 4 7 e 2 x - 5 = 58. e ln 6 0.6 f ln 6 0.6 L 2.99. Discovery Use properties of logarithms to show that the solution in Example 4 can be expressed as ln 1600 ln A 5 16 B .
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Section 3.4 Exponential and Logarithmic Equations 427 Technology Graphic Connections Shown below is the graph of There are two one at 0 and one at approximately 1.10. These intercepts verify our algebraic solution. [ 3, 3, 1] by [ 1, 3, 1] x -intercepts, y = e 2 x - 4 e x + 3. Solving an Exponential Equation Solve: Solution The given equation is quadratic in form. If the equation can be expressed as Because this equation can be solved by factoring, we factor to isolate the exponential term. This is the given equation. Factor on the left. Notice that if Set each factor equal to 0. Solve for Take the natural logarithm on both sides of the first equation. The equation on the right can be solved by inspection. The solution set is The solutions are 0 and ln 3, which is approximately 1.10. Check Point 5 Solve: Find the solution set and then use a calculator to obtain a decimal approximation to two decimal places, if necessary, for the solutions.
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  • Fall '16
  • Kirkpatrick
  • Physics, pH, Natural logarithm, Logarithm

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