# Looking for ideas on this.

• May 23rd 2012, 01:45 PM
Jimbo32
Looking for ideas on this.
Hi, I am currently struggling with Logarithms and exponentials in pre calc. the two in particular are f(x)=-2+logbase 1/2(x+2) sketch the graph and identify domain range x/Y intercept and end behaivor. and

In mediaeval times there were 10,000 people living in a city that was struck by a plague. the number of people living after 6 days was 8500, and after 12 was 7225 explain how you determine it is exponential and fill in a table and find an equation/common ratio.

Thanks a lot for any ideas or help for on where to start.
• May 23rd 2012, 02:09 PM
skeeter
Re: Looking for ideas on this.
$\displaystyle f(x) = -2 + \log_{\frac{1}{2}}(x+2)$

note that you can change the base ...

$\displaystyle \log_{\frac{1}{2}}(x+2) = \frac{\log_2(x+2)}{\log_2\left(\frac{1}{2}\right)} = -\log_2(x+2)$

$\displaystyle f(x) = -2 - \log_2(x+2) = -[2 + \log_2(x+2)]$

domain ...

$\displaystyle x+2 > 0 \implies x > -2$

range would be all reals

y-intercept ... evaluate $\displaystyle f(0)$

x-intercept ... $\displaystyle f(x) = 0 \implies \log_2(x+2) = -2$

graph will be the function $\displaystyle y = 2 + \log_2(x+2)$ reflected over the x-axis

second problem ... $\displaystyle \frac{8500}{10000} = 0.85$ , what is the value of $\displaystyle \frac{7225}{8500}$ ? does that tell you something?
• May 23rd 2012, 02:24 PM
Jimbo32
Re: Looking for ideas on this.
and after achieving a common factor getting my equation would be something like your common factor raised^x power? wasn't very good at it.
• May 23rd 2012, 02:33 PM
skeeter
Re: Looking for ideas on this.
Quote:

Originally Posted by Jimbo32
and after achieving a common factor getting my equation would be something like your common factor raised^x power? wasn't very good at it.

so, let's cut the self-deprecation ... what is the equation you came up with?
• May 23rd 2012, 04:36 PM
Jimbo32
Re: Looking for ideas on this.
I got something like 2*.85^X but I dont know if i did it right. I took the difference in the days 12/6 which was two and multiplied that by my common factor and raised it to X because that was changing. though that doesnt seem right.
• May 23rd 2012, 05:49 PM
skeeter
Re: Looking for ideas on this.
Quote:

Originally Posted by Jimbo32
I got something like 2*.85^X but I dont know if i did it right. I took the difference in the days 12/6 which was two and multiplied that by my common factor and raised it to X because that was changing. though that doesnt seem right.

every six days, the population decreases by 15%

population as a function of time , t , in days ...

$\displaystyle P(t) = 10000(0.85)^{\frac{t}{6}}$

... you really need to have a sit-down with your instructor or a real live tutor.