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
implies for example 3^x=3^2x-5 therefore x=2x-5
and when it comes to logs with the same base
implies x=7Solving Exponential Equations
The first thing to recognize is that you can take either an algebraic or logarithmic approach when solving this equation.
first take the natural log of both sides
natural log of e is simply e, so your next step becomesx=ln17 which you plug into your calculator to find that
x=2.83 (remember to round to the nearest hundredth..or else)
first rewrite as the equivalent logarithmic, then getx=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
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