Exponents and logarithms, multiple pages

Problem 1:
The magnitude $R$ of an earthquake of intensity $I$ , on the Richter scale, is:
$R=\log \frac{I}{I_0}$
Where $I_0$ is a certain minimum intensity.
For an intensity of an earthquake of $100000I_0$ , what is the magnitude $R$ ?

Solution:

$R=\log \frac{I}{I_0}$
$R=\log \frac{100000I_0}{I_0}$
$R=\log 10^{5}$
$R=5log 10$
But $log 10 =1$

Finally:
The magnitude $R=5$

Answer: $Magnitude=5$

Problem 2:
A certain population is growing continuously at the rate of 7% per year.
What time approximately it takes the population to double its current value?

Solution:

We use:
$P=P_0e^{kt}$ with $k=0.07$
For it to double we have:
$2P_0=P_0e^{0.07t}$
$2=e^{0.07t}$
$\ln 2=0.07t$ because $\ln e=1$
$t=\frac{\ln 2}{0.07}$
$t=9.9$ years.

Answer:
It doubles in $9.9 \; years$

Problem 3:
The interest is compounded continuously at the rate of 4% per year. In how many years,
approximately, a deposit of 7000 dollars willgrow to 22000 dollars?

Solution:
We use
$A(t)=A_0e^{rt}$ with $A(t)=22000$ , $A_0=7000$ and $r=0.04$

$22000=7000e^{0.04t}$
$\frac{22000}{7000}=e^{0.04t}$
$\frac{22}{7}=e^{0.04t}$
$\ln 22- \ln 7=0.04t$
$t \approx 29$
Answer: Time needed: 29 years

Problem 4:
A certain substance decays according to the formula: $A(t)=A_0b^{-t}$
What is the value of $b$ if the half life of the substance is 12 days?

Solution:

Given the formula:
$A(t)=A_0b^{-t}$
Half life is $\frac{1}{2}A_0$
We get:
$\frac{1}{2}A_0=A_0b^{-12}$
$\frac{1}{2}=b^{-12}$
Many options from here:
We notice the exponents:
$\frac{1}{2}=\frac{1}{b^{12}}$

Or simply:
$2=b^{12}$
$b=\sqrt[12]{2}$
$b=1.05946$

Answer: $b=1.05946$

Problem 5:
The number of bacteria in a culture is given by $N(t)=3(4^t)$ where $t$ is in hours.
If $N(t)$ is in tens of thousands:
What is the number of bacteria after 150 minutes?

Solution:

$t=2.5$ because 150 minutes equal 2.5 hours.

$N(2.5)=3(4^{2.5})$
$N(2.5)=96$
Finally:
Number of bacteria: $96\times 10000=960000$

Answer: Total number of bacteria after 150 minutes: 960000

Problem 6:
Ben Franklin left 4000 dollars for the city of Philadelphia with instructions that it should be used after 200 years.
Based on a continuous compounding, it was found to be about 2 millions dollars.
Find the inerest rate.

Solution:
$A(t)=A_0e^{kt}$
$2000000=4000e^{200k}$
$500=e^{200k}$
$\ln 500=200k$
$k=\frac{\ln 500}{200}$
$k=0.031$

Answer: The rate is: $3.1\%$

Problem 7:
The formula $T=-8310\ln x$ is sometimes used in carbon 14 dating of some bone fossils.
$T$ being the age in years and $x$ is the percentage of carbon 14 remaining, expressed in decimals.
What is the age of a bone having 6% of carbon 14?
If a fossil is 15000 years old, what is the percentage of carbon 14present in the fossil?

Solution:
If $x=0.06$ :
$T=-8310 \ln 0.06$
$T=23379$ years.

If $T=15000$ :
$-8310 \ln x=15000$
$\ln x=\frac{15000}{-8310}$
$\ln x=-1.80505415$
$x=e^{-1.80505415}$
$x=0.16447$

Answer: Carbon 14 present: 16.447%

Problem 8:
In 1971 the minimum wage in the United States was $1.60$ dollars per hour.
If the rate of inflation is 5% per year, find the equivalent minimum wage in 2015.

Solution:
$A(t)=A_0e^{kt}$ , $k=0.05$ and $t=44$
$A(t)=1.60e^{0.05\times 44}$
$A(t)=14.45$

Answer: Minimum wage in 2015: $14.45\; dollars$

Problem 9:
The population $N(t)$ , in millions, of the United States $t$ years after 1980 may be appromated by the formula
$N(t)=227e^{0.007t}$ . When will the population be double of what it was 1980?

Solution:
$N(t)=227e^{0.007t}$
In 1980:
$N(0)=227e^{0.007\times 0}$
$N(0)=227$ in millions
To double:
$2N(0)=N(0)e^{0.007t}$
$2=e^{0.007t}$
$\ln 2=0.007t$ , because $\ln e=1$
$t=\frac{-\ln 2}{0.007}$
$t=99$
The year it doubles is $1980+99=2079$

Answer: It will be double in 2079

Problem 10:
The population $N(t)$ , in millions, of India $t$ years after1985 may be approximated by:
$N(t)=762e^{0.022t}$
When will the population be 1.5 billion?

Solution:

$N(t)=762e^{0.022t}$
Initially, in 1985:
$N(0)=762$ in millions.
$N(t)=1500$ millions
$1500=762e^{0.022t}$
$\frac{1500}{762}=e^{0.022t}$
$\ln \frac{1500}{762}=0.022t$
$t=\frac{\ln 1500- \ln 762}{0.022}$
$t \approx 31$ years
The year it will be 1.5 billion is: $1985+31=2016$

Answer: It will be 1.5 billion in 2016.

Problem 11:

Pareto’slaw for capitalist countries states that the relationship between annual income
$x$ and the number $y$ of individuals whose income exceeds $x$ is:
$\log y=\log b- k\log x$
Where b and k are positive constants. Solve the equation for $y$