# Thread: Need solutions for some easy calc problems

1. ## Need solutions for some easy calc problems

thanks

=============================

It makes it more difficult for someone else to benefit from the
solutions.

2. Originally Posted by JohnSena

2. Use logrithms to solve the quation 25(1.06)^x=100 for x
?
You have,
$25(1.06)^x=100$
Divide sides by 25,
$(1.06)^x=4$
Take, logarithm
$\log (1.06)^x=\log 4$
Thus, the exponent "comes down"
$x\log (1.06)=\log 4$
Thus,
$x=\frac{\log 4}{\log 1.06}\approx 23.8$

3. Originally Posted by JohnSena
3. Solve for x using logs. a) 2e^4x=8e^6x b) 3^x+3=e^7x
a)You have,
$2e^{4x}=8e^{6x}$
Divide by 2,
$e^{4x}=4e^{6x}$
Divide by $e^{4x}$ thus,
$1=4e^{6x-4x}=4e^{2x}$
Divide by four,
$1/4=e^{2x}$
Take natural logarithm of both sides,
$\ln .25=2x$
Thus,
$x=.5\ln .25\approx -.69$

4. damn that was quick thanks

still confused on how to figure out part b and the other problems though, if anybody could help with those that'd be awesome

5. Originally Posted by JohnSena
1. A town has 1000 people initially. Find the formula for the population of the town, P, in terms of the number of years, t. a) The town grows at an annual rate of 8% a year. b) The town grows by 70 people a year
a) the rate of growth of population is proportional to the existing population.
So the population satisfies this differential equation:

$
\frac{dp}{dt}=\alpha p(t)
$
,

which has solution:

$
p(t)=p(0) e^{\alpha t}
$
.

Now we are told that the population grows by $8\%$ per year, so
after one year the population is $1.08$ times its original value so:

$
p(1)=1.08 p(0)=e^{\alpha}p(0)
$
,

so: $\alpha=\log(1.08)$ (thats a natural log by the way).

hence:

$
p(t)=p(0) e^{\log(1.08) t}=1000 e^{\log{1.08}t}
$
.

6. Originally Posted by JohnSena
1. A town has 1000 people initially. Find the formula for the population of the town, P, in terms of the number of years, t. a) The town grows at an annual rate of 8% a year. b) The town grows by 70 people a year
b) The rate of population growth is a constant so the population satisfies
the differential equation:

$
\frac{dp}{dt}=70
$
,

which has solution:

$
p(t)=70t+p(0)
$
,

plugging in the given initial population gives:

$
p(t)=70t+1000
$
.

RonL

7. Originally Posted by JohnSena
4. If the size of a bacteria colony doubles in 8 hours, how long will it take for the number of bacteria to be 5 times the original amount?
Missing assumption needed to solve this is that the rate of growth of the
population is proportional to the existing population. Hence the population
satisfies the diffrenetial equation:

$
\frac{dp}{dt}=\alpha p
$
,

which has solution:

$
p(t)=p(0) e^{\alpha t}
$

Now we are told that the population doubles in 8 hours so:

$
p(8)=2p(0)=p(0) e^{8\alpha}
$
,

so: $2=e^{8\alpha}$, or: $\alpha=\frac{\log(2)}{8}
$
.

Therefore:

$
p(t)=p(0)e^{\frac{\log(2)}{8}t}
$
.

Now at time $t_5$ the population is 5 times its original value so:

$
p(t_5)=5p(0)=p(0)e^{\frac{\log(2)}{8}t_5}
$
,

so:

$
5=e^{\frac{\log(2)}{8}t_5}
$

or laking logs:

$
\log(5)=\frac{\log(2)}{8}t_5
$
,

or:

$
t_5=\frac{8\log(5)}{\log(2)}
$

RonL

8. Originally Posted by JohnSena
thanks