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This is the given equation.Add 3 to both sides.Isolate the exponential factor by dividing bothsides 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 60.6L2.99lnex=x0.6x=ln 6ln e0.6x=ln 6e0.6x=640e0.6x=24040e0.6x-3=237e0.6x.40e0.6x-3=237.EXAMPLE 3Thus, the solution of the equation is Try checking this approximatesolution in the original equation to verify thatis the solution set.Check Point3Solve:Find the solution set and then use acalculator to obtain a decimal approximation to two decimal places for the solution.Solving an Exponential EquationSolve:SolutionBecause each exponential expression is isolated on one side of theequation, 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 variableexponents to the front: Use the distributive property to distributeln 5 and ln 4 to both terms in parentheses.Collect variable terms involving on the left by subtracting and adding 2 ln 5on 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 Point4Solve:Find the solution set and then use acalculator to obtain a decimal approximation to two decimal places for the solution.32x-1=7x+1.-6.34.e2 ln 5+3 ln 4ln 5-2 ln 4f.ln 5-2 ln 4.xx=2 ln 5+3 ln 4ln 5-2 ln 4xx1ln 5-2 ln 42=2 ln 5+3 ln 42xln 4xxln 5-2xln 4=2 ln 5+3 ln 4xln 5-2 ln 5=2xln 4+3 ln 4lnbx=xlnb.(x-2) ln 5=(2x+3) ln 4Be sure to insert parentheses around the binomials.Remember that ln 5 and ln 4 are constants, not variables.ln 5x-2=ln 42x+35x-2=42x+35x-2=42x+3.EXAMPLE 47e2x-5=58.eln 60.6fln 60.6L2.99.DiscoveryUse properties of logarithms to showthat the solution in Example 4 can beexpressed asln 1600lnA516B.
Section 3.4Exponential and Logarithmic Equations427TechnologyGraphic ConnectionsShown below is the graph ofThere are twoone at 0 and one atapproximately 1.10. These interceptsverify our algebraic solution.[−3, 3, 1] by [−1, 3, 1]x-intercepts,y=e2x-4ex+3.Solving an Exponential EquationSolve:SolutionThe given equation is quadratic in form. If the equation can beexpressed as Because this equation can be solved by factoring, wefactor 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 sidesof 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 approximately1.10.Check Point5Solve:Find the solution set and then use acalculator to obtain a decimal approximation to two decimal places, if necessary,for the solutions.