Algebra/Trig Review - Pauls Online Math Notes - Lamar University

Algebra/Trig Review⎛ x ⎞2ln ⎜ + 3⎟=−7⎝7⎠⎛ x ⎞ 7ln ⎜ + 3⎟=−⎝7 ⎠ 2e⎛ x ⎞ln 37⎜ + ⎟ −⎝7 ⎠ 2= e7x −2+ 3 = e7Now, solve this.7x −2+ 3 = e77x−2=− 3 + e77⎛ − ⎞2x = 7⎜− 3+e ⎟⎝ ⎠x = -20.78861832xI’ll leave it to you to check that + 3 will be positive upon plugging7x = − 20.78861832 into it and so we’ve got the solution to the equation.3. ( x) ( x)2ln −ln 1− = 2SolutionThis one is a little different from the previous two. There are two logarithms in theproblem. All we need to do is use Properties 5 – 7 from the Logarithm Propertiessection to simplify things into a single logarithm then we can proceed as we did in theprevious two problems.The first step is to get coefficients of one in front of both logs.2ln x −ln 1− x = 2( ) ( )( x) ( x)ln −ln 1− = 2Now, use Property 6 from the Logarithm Properties section to combine into thefollowing log.⎛ x ⎞ln ⎜ ⎟ = 2⎝1−x ⎠Finally, exponentiate both sides and solve.© 2006 Paul Dawkins 96http://tutorial.math.lamar.edu/terms.aspx