logs: how to graph log_(1/2) x - 3 (don't understand logs)

[jkwehalkrh]

how would you graph the function f(x)=log_1/2x-3
i dont understand logs so im guessing the parent function is log_1/2 ?
and the asymptote is going to x=3 i think

 

[pka]

how would you graph the function f(x)=log_1/2x-3
i dont understand logs so im guessing the parent function is log_1/2 ?
and the asymptote is going to x=3 i think

\(\displaystyle \log_{.5}(x-3)=\dfrac{\log(x-3)}{\log(.5)}=\dfrac{\log(x-3)}{-\log(2)}=\dfrac{-1}{\log(2)}\log(x-3)\).

GRAPH

 

[greg1313]

\(\displaystyle \log_{.5}(x-3)=\dfrac{\log(x-3)}{\log(.5)}=\dfrac{\log(x-3)}{-\log(2)}=\dfrac{-1}{\log(2)}\log(x-3)\).

\(\displaystyle \log_{.5}(x-3) \ = \ \dfrac{\log(x-3)}{\log(.5)} \ = \ \dfrac{\log(x - 3)}{\log(2^{-1})}\ = \ \dfrac{\log(x-3)}{-\log(2)} \ =\ \dfrac{-1}{\log(2)}\log(x-3)\).




The inclusion of that middle expression justifies your third expression, and it doesn't "lose the student around the corner."

 

[stapel]

how would you graph the function f(x)=log_1/2x-3
i dont understand logs so im guessing the parent function is log_1/2 ?

I'm not sure what you mean by "the parent function"...? And what do you mean when you say that you "don't understand logs"? Were they not covered in class? Or are you just having difficulty with one particular area or application...?

 

[pka]

The inclusion of that middle expression justifies your third expression, and it doesn't "lose the student around the corner."

You don't have much teaching experience do you.
That was the whole point: throw a curve. Make the student justify the step.