Logarithmic Function

[joeyjon]

Completely stumped. I did very well in this chapter of College Algebra, but this one got me. I don't even know where to begin...

log (x²) = (log x)²

The back of the book gives the solution set of { 1, 100 }.

I hope someone can explain this to me.

Thank you!

 

[tkhunny]

Use your log rules.

\(\displaystyle 2\cdot log(x) = log(x) \cdot log(x)\)

\(\displaystyle 2\cdot log(x) - log(x) \cdot log(x) = 0\)

\(\displaystyle [2 - log(x)][log(x)] = 0\)

you studied factoring for a reason. Thsi is a bad time to forget.

 

[joeyjon]

Thank you.

I need more practice it seems. Even after you showed me what you did (which I did understand the dividing becoming subtracting and the multiplying becoming addition), it took me awhile to recognize 2 separate equations, 2=log (x) and log x = 0.

Then I just guessed with buttons on my calculator for the 2 = log (x). I didn't even know by looking at that that it is a "common logarithm" with 10. I obviously knew it wasn't a natural logarithm, but I get confused with the common logarithm and just pressing the log button on my calculator.

This is one of those last problems at the end of the section, so there aren't anymore for me to work. The rest of the problems in the section were just easier.

I guess I'll go back and redo the chapter on logarithms because I don't feel very confidant with them.

Anyway, thanks for your help. I don't have a teacher. I just bought an algebra book and am trying to figure it out.

 

[tkhunny]

There is historical confusion with "log(x)". Before you get to calculus, it generally mean "base 10" or "common". During calculus, it changes to "base e" or "natural". It is very interesting that calculators DON'T make the switch, but retain the pre-calculs definitions. This is only because of potential ambiguity.