Question 35519
{{{log y =log(0.5x-3) + log 2}}}


By the  first law of logarithms, the sum of two logs is the log of the product.  See my Lesson Plans in algebra.com on logarithms if you need help with this.

{{{log y = log 2(0.5x -3) }}}
{{{ log y = log (x -6) }}}


Now, if the log of THIS = the log of THAT, then THIS = THAT, so 
{{{ y = x-6}}} BUT you must remember the restriction that you can never have the log of a negative, so you are restricted to values of x > 6.  Do you see why they (and we!) said that?


So this is a straight line starting at the point (6,0) with an open circle to indicate that this point is NOT included.  From that point, draw a straight line with a slope of 1 (Rise=1, and Run = 1).  I'll draw it on this grapher, but I don't know exactly how to limit the graph as I have described.  I'll have to draw the whole line, and then I'm asking YOU to make an open circle at (6,0) and from this point go UP the graph.  Everything below the x-axis should NOT be there.


{{{graph(300,300, -3, 10, -3, 10, x-6) }}}


R^2 at SCC