simple geometric sequence question

**differentiate** · November 27th 2009, 07:13 PM

The number of people attending the yearly Abletown show is rising by 5% per annum, and the number attending the yearly Bush Creek Show is falling by 5% per annum. in the first year under consideration, 5000 people attended both shows
Show that the number attending the Abletown show first exceeds ten times the number attending the Bush Creek Show in the 25th year.

**Soroban** · November 27th 2009, 08:05 PM

Hello, differentiate!

I get a different answer . . .

The number of people attending the yearly Abletown show is rising by 5% per year,
and the number attending the yearly Bush Creek show is falling by 5% per year.
In the first year under consideration, 5000 people attended both shows.

Show that the number attending the Abletown show first exceeds ten times
the number attending the Bush Creek Show in the 25th year.

The number attending the Abletown show in year $n$ is: . $A \:=\:5000(1.05)^n$

The number attending the Bush Creek show in year $n$ is: . $B \:=\:5000(0.95)^n$

When is $A$ greater than 10 times $B$ ?

. . $5000(1.05)^n \;>\;10\cdot5000(0.95)^n \quad\Rightarrow\quad (1.05)^n \;>\;10(0.95)^n$

. . $\frac{(1.05)^n}{(0.95)^n} \;>\:10 \quad\Rightarrow\quad \left(\frac{1.05}{0.95}\right)^n \;>\;10$

Take logs: . $\ln\left(\frac{1.05}{0.95}\right)^n \;>\;\ln(10) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.05}{0.95}\right) \;>\;\ln(10)$

Therefore: . $n \;=\;\frac{\ln(10)} {\ln\left(\frac{10.05}{0.95}\right)} \;=\;23.00664991$

The event will occur in the ${\color{blue}24^{th}}$ year.

**differentiate** · November 29th 2009, 09:36 PM

thanks it was the same answer i got too

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