# Logarithmic functions?

• May 25th 2011, 12:16 PM
homeylova223
Logarithmic functions?
Graph each equation or inequality

1.y= log 5(x-1) This is what I did
y+1= log 5 x
5^y+1=x Would this be right of setting up the graph so that if y=0 x=5?

2. y=2 log 2 x I am slightly confused on how to start this
• May 25th 2011, 12:23 PM
TheEmptySet
Quote:

Originally Posted by homeylova223
Graph each equation or inequality

1.y= log 5(x-1) This is what I did
y+1= log 5 x
5^y+1=x Would this be right of setting up the graph so that if y=0 x=5?

2. y=2 log 2 x I am slightly confused on how to start this

I just want to be clear about your notation do you mean

$\displaystyle y=\log_{5}(x-1) \text{ or } y=\log_{5}(x)-1$
• May 25th 2011, 12:28 PM
homeylova223
I mean the former the one where both x and negative one are in the parenthesis.
• May 25th 2011, 12:38 PM
TheEmptySet
First you can graph in the method that you started above but you don't have to.

$\displaystyle y=\log_{5}(x-1) \iff 5^y=x-1 \iff x=5^y+1$

You can then pick y values to make a table. Note that the x values get large really quickly.

Or you can make the table directly from the function. First we know that the log of zero is undefined so when

$\displaystyle x-1=0 \iff x=1$

so there is a vertical asymptote when x=1.

Also you should know that

$\displaystyle \log_{a}(1)=0$

for any

$\displaystyle a > 0, \quad a \ne 1$

Now we can just pick powers of the bases to get

$\displaystyle \log_{5}\left(5^{n} \right)=n$

can you finish from here?
• May 25th 2011, 01:36 PM
homeylova223
For my second question

y=2 log 2 x Would I do this

2^y=2x Then would I just divide by 2?
• May 25th 2011, 03:08 PM
skeeter
Quote:

Originally Posted by homeylova223
For my second question

y=2 log 2 x Would I do this

2^y=2x Then would I just divide by 2?

if you meant ...

$\displaystyle y = 2\log_2{x}$

then,

$\displaystyle 2^y = x^2$