Precalculus 1/13/09

Student quote of the day " e is hiding....." Kush

Today we learned how to solve equations that involved logarithms. Here are a couple examples chosen by the 5th period students...

Example #1 An equation involving natural logarithms




subtract 4 from both sides

divide both sides by 3

write in exponential form "e is hiding"

solve for x

Example #2 equation involving multiple logarithms




use log property (log A+log B) = log(A*B)

write in exponential form " 2 is not hiding"

simplify

get equal to zero

factor

x=5 and x = -1 would seem to be the answers, but -1 does not work (it's called an extraneous solution) because it makes the argument(the (x-1) or x-3) spot in the original equation) negative. So the answer is x = 5

Helpful hint: Always get log or ln by itself before turning an equation into exponential form.

You can find other notes at Mr. Corn's web page here