Question 274271
{{{y=3-log(2,(2x+5))}}}
I'm not sure what you mean by "key points". The only ones I can think of would be the x and y intercepts, if any. The x-intercept is the point(s) where a graph crosses/touches/intersects the x-axis. Since all the points on the x-axis have a y coordinate of zero, that is how you find them. You make y zero and solve for x:
{{{0=3-log(2,(2x+5))}}}
Adding the log to both sides we get:
{{{log(2,(2x+5)) = 3}}}
Rewriting this in exponential form we get:
{{{2x+5 = 2^3}}}
which simplifies to:
{{{2x+5 = 8}}}
Subtracting 5 we get:
{{{2x = 3}}}
Dividing by 2 we get:
{{{x = 3/2}}}
So the x-intercept is (3/2, 0)<br>
For the y-intercept we make x zero:
{{{y=3-log(2,(2(0)+5))}}}
which simplifies to:
{{{y=3-log(2,(5))}}}
Since our calculators can't "do" base 2 logarithms we need to change the base of the logarithm using the conversion formula, {{{log(a, (p)) = log(b, (p))/log(b, (a))}}}, to convert to a base our calculator "knows" (like base 10 or base e (ln)). I'll use base 10:
{{{y=3-log((5))/log((2))}}}
{{{y=3-0.6989700043360188/0.3010299956639812}}}
{{{y=3-2.3219280948873623}}}
{{{y = 0.6780719051126377}}} which I'll round off to 0.7. So the y-intercept is near (0, 0.7)<br>
Another item which can be helpful when graphing is the vertical asymptote. The arguments of all logarithms must be positive. So the domain of any equation like yours must have a domain which makes your argument (2x+5) positive. In other "words":
{{{2x+5 > 0}}}
Solving this we get:
{{{2x > -5}}}
{{{x > -5/2}}}
So the domain is all numbers greater than -5/2.<br>
We find the domain because the "edge" of the domain will be a vertical asymptote. Just make an equation out of the domain:
x = -5/2
and you have your vertical asymptote!<br>
Two points and an asymptote are not much to go on. You may need other points. To find other points just pick numbers for x or y and use the equation to find the value of other variable (just like we did when we picked 0 for x and found the y-intercept and when we picked 0 for y and found the x-intercept). I'll leave it up to you to find more points.<br>
Algebra.com's graphing feature is not perfect but here's how the graph should look (roughly):
{{{graph(400, 400, -3, 17, -10, 10, 3-log(2,2x+5))}}}