# Math Help - [SOLVED] Help to solve x and find population

1. ## [SOLVED] Help to solve x and find population

Okay my boyfriend and I worked on these two problems last night and couldn't figure it out.

Problem 1:

Solve for :

which I got: .8348214
and Dj said it was undefined.
which both were wrong.
Problem 2:

The rat population in a major metropolitan city is given by the formula where is measured in years since 1994 and is measured in millions.

What was the actual rat population in 1994 ? (not "in millions")
What is the rat population going to be in the year 2006 ?

2. Since t is the time in years since 1994, the time t=0 corresponds to the year 1994. Therefore the rat population in 1994 was equal to $n(0)=29 e^{0.015*0}=29*1=29$.

Now 2006=1994+12 and so the rat population in 2006 will be equal to $n(12)=29 e^{0.015*12}=29*e^{0.18}$.

3. Rewrite the first one as:

$\frac{6^{3x}}{6^{9}}=\frac{7^{6x}}{7^7}$

Use log laws:

$log(6^{3x})-9log(6)=log(7^{6x})-7log(7)$

$log(216^{x})-log(117649^{x})=9log(6)-7log(7)$

$x(log(216)-log(117649))=9log(6)-7log(7)$

$x=\frac{9log(6)-7log(7)}{log(216)-log(117649)}=\frac{9log(6)-7log(7)}{3log(6)-6log(7)}\approx -.3975225...$

Or some other form you may like. It's just a matter of writing it differently using the exponents and log laws.

4. Originally Posted by nmatthies1
Since t is the time in years since 1994, the time t=0 corresponds to the year 1994. Therefore the rat population in 1994 was equal to $n(0)=29 e^{0.015*0}=29*1=29$.
okay, I already tried 29 and it was wrong.

Originally Posted by nmatthies1
Now 2006=1994+12 and so the rat population in 2006 will be equal to $n(12)=29 e^{0.015*12}=29*e^{0.18}$.

this one I got: 34.71930353
and it also was wrong

5. Originally Posted by lsnyder
okay, I already tried 29 and it was wrong.

this one I got: 34.71930353
and it also was wrong
That is the correct solution. Almost 35 million rats

6. Originally Posted by lsnyder
Okay my boyfriend and I worked on these two problems last night and couldn't figure it out.

Problem 1:

Solve for :

which I got: .8348214
and Dj said it was undefined.
which both were wrong.
1. Take logs of both sides. I will pick base e but it really doesn't matter and remember that $ln{(a^k)} = k\ln(a)$

$(3x-9)ln(6) = (6x-7)ln(7)$

$3x-9 = \frac{ln(7)}{ln(6}(6x-7) = \frac{6x\ln(7)}{ln(6)} - \frac{7ln(7)}{ln(6)}$
$
3x - \frac{6x\ln(7)}{ln(6)} = 9 - \frac{7ln(7)}{ln(6)}$

$x(3-\frac{6ln(7)}{ln(6)}) = 9 - \frac{7ln(7)}{ln(6)}$

$x = \frac{9 - \frac{7ln(7)}{ln(6)}}{3-\frac{6ln(7)}{ln(6)}} =$

7. Originally Posted by galactus
That is the correct solution. Almost 35 million rats

well my homework is telling me thats wrong everytime type it in.
and the beginning formula is talking about since 1994 (thats what I understand from the problem). They want to know the population of 1994, not since. But I could be wrong, that's just what I took from the problem.

8. Originally Posted by lsnyder
well my homework is telling me thats wrong everytime type it in.
and the beginning formula is talking about since 1994 (thats what I understand from the problem). They want to know the population of 1994, not since. But I could be wrong, that's just what I took from the problem.
I don't see any other way of solving this problem. Unless you haven't stated it correctly.

9. here is problem 2 again
(copied and pasted)

The rat population in a major metropolitan city is given by the formula where is measured in years since 1994 and is measured in millions.
What was the actual rat population in 1994 ? (not "in millions") What is the rat population going to be in the year 2006 ?

10. It says, "not in millions". The only thing I see is to type in the 34.7

Unless they want it typed in as 34,719,304 rats.

11. okay i got it.
i didnt realize it wanted written ver specifically.
thank you for all your help