Solving Exponential and Logarithmic Equations- Sec. 3.4

        Friday's lesson consisted of learning the procedures used in solving exponential and logarithmic equations. Before we delved into that, though, we reviewed what it meant for a function to be "one-to-one."

One-to-one:

Where    implies  

for example    3^x=3^2x-5 therefore x=2x-5

and when it comes to logs with the same base implies x=7

Solving Exponential Equations

Example one: Solve for x when

The first thing to recognize is that you can take either an algebraic or logarithmic approach when solving this equation.

If solving this equation algebraically:first take the natural log of both sides
                                                             natural log of e is simply e, so your next step becomes
                                                               x=ln17         which you plug into your calculator to find that
                                                               x=2.83        (remember to round to the nearest hundredth..or else)

If solving this equation logarithmically: first rewrite as the equivalent logarithmic, then get
                                                               x=ln17          (since log base e is representative of ln)
                                                               x=2.83         "  "

Example two: Using log algebraically:  2 + 5^(x-3) = 12  
                                                                 5^x-3=10           subtract 2
                                                           log[5^(x-3)]=log10   take log of both sides
                                                                 (x-3)log5=1        move x-3 out front and log base10 of 10=1
                                                                 x-3=1/log5         divide by log 5
                                                                x=3+(1/log5)       add 3 to each side
                                                                    x=4.43            round to nearest hundredth

                                                                                         (remember that you can easily 
                                                                                               your answer by plugging it
                                                                                                into the original equation)    

The final example Mr.Wilhelm showed us was one of an exponential equation in need of solving that looked eerily similar to a quadratic equation.

Example three:            e^(2x) - 3e^x + 2 = 0         first let u=e^x
                                           u^2 - 3u +2 = 0
                                            (u-2) (u-1) = 0    
                                         e^x = 2    e^x = 1           substitute e^x back in
                                 ln e^x = ln 2     ln e^x = ln 1    take the natural log of all four sides
                                   x = ln 2                 x=0          ln 1 implies e^x must equal 1 so therefore x is 0 and
                                   x= .69                                    

Homework: Section 3.4 problems 3,5,12,14,16-40 multiples of 4, 46, 47, 50