# Graphing Logarithms

• Apr 25th 2008, 10:01 AM
trainman238
Graphing Logarithms
Okay, so I know that for graphing logs by hand, you can rewrite it as an exponential... for example, log base 5 of x can be rewritten as 5^? = x...
But, I was just wanting to be sure... I have log base 3 of (x-5) + 3... so, would the exponential form be 3^? = (x-5) + 3?.... I am pretty sure it is... Because then I have to fill out an x y chart to go with it and then I have to graph it on paper.... So, yeah... just wanting to make sure...
• Apr 25th 2008, 10:33 AM
Jhevon
Quote:

Originally Posted by trainman238
Okay, so I know that for graphing logs by hand, you can rewrite it as an exponential... for example, log base 5 of x can be rewritten as 5^? = x...
But, I was just wanting to be sure... I have log base 3 of (x-5) + 3... so, would the exponential form be 3^? = (x-5) + 3?.... I am pretty sure it is... Because then I have to fill out an x y chart to go with it and then I have to graph it on paper.... So, yeah... just wanting to make sure...

Graphing log graphs are no where near that complicated. Is this how you have to do it?

You can graph any log graph by shifting the general form of the log graph. See below:

All graphs of the form $y = \log_a x$ looks like that. I do not care what the base $a$ is. They all pass through (1,0), shape like a square root graph and are asymptotic to the y-axis. To get the graph $y = \log_3 (x - 5) + 3$, you take the graph you see below, shift it 5 units to the right (so the asymptote shifts as well, 5 units, and ends up at the line x = 5). Then shift that graph 3 units up (because of the + 3). Now you would simply solve for the x- and y-intercepts. There are no y-intercepts as you can probably tell. to find the x-intercept, set y = 0 and solve for x. then you will know where you graph touches the x-axis
• Apr 25th 2008, 11:32 AM
trainman238
thanks man... that REALLY helped a lot..
• Apr 25th 2008, 11:41 AM
trainman238
thanks man... that REALLY helped a lot..