Expert: Paul Klarreich - 11/18/2011

Question
Hi Paul, I'm in a grade 12 math course and we are studying logarithmic equations. I can't quite figure out the last question on my assignment. The instructions say "Solve for the unknown in each of the following questions. Round the to nearest hundredth. Evaluate logarithms at the very end.". The equation is log(x^2 - 2x +3) = 3. I really have no idea how to begin. I have tried setting it up like so: logx^2 - log2x + log3 = 3, but that didn't make sense. I'm at a loss and would greatly appreciate your help.

Thanks in advance!

-Michelle

Answer
Questioner: Michelle
Country: Prince Edward Island, Canada
Category: Advanced Math
Private: No
Subject: Solving a logarithmic equation
Question: Hi Paul, I'm in a grade 12 math course and we are studying logarithmic equations. I can't quite figure out the last question on my assignment. The instructions say "Solve for the unknown in each of the following questions. Round the to nearest hundredth. Evaluate logarithms at the very end.". The equation is log(x^2 - 2x +3) = 3. I really have no idea how to begin. I have tried setting it up like so: logx^2 - log2x + log3 = 3, but that didn't make sense. I'm at a loss and would greatly appreciate your help.

Thanks in advance!

-Michelle
..........................................
You are assuming that:

log(x^2 - 2x +3) = log x^2 - log 2x + log 3 ???

YECCCHH!!!!!

You must use the KNOWN properties of the log function, such as:

log AB = log A + log B

Now, about your example:

log(x^2 - 2x +3) = 3

Generally, you are supposed to know the base of the log function;

If you write  log[7](...), the base is 7.
If you write  ln(...), the base is e. [You will learn about that in calculus.]
If you write  log(...), the base is 10.[You will learn about that in 11th year math -- Oh, yes, that was last year.]

Anyway, I shall assume log meant base 10.  Soooooo...

log A = B  means  A = 10^B

log(x^2 - 2x +3) = 3  means

x^2 - 2x +3 = 10^3

x^2 - 2x +3 = 1000

x^2 - 2x - 997 = 0

That's a quadratic.  Over to you.