1. find the annual interest rate. (round your answer to the nearest whole number.)\nprincipal $3000 balance…

1.

Explanation:

Step1: Recall compound - interest formula

The compound - interest formula is $A = P(1+\frac{r}{n})^{nt}$, where $A$ is the balance, $P$ is the principal, $r$ is the annual interest rate (in decimal), $n$ is the number of times compounded per year, and $t$ is the number of years. Here, $P = 3000$, $A=38754.85$, $n = 4$ (quarterly compounding), and $t = 20$. So, $38754.85=3000(1 +\frac{r}{4})^{4\times20}$.

Step2: Isolate the exponential term

First, divide both sides by 3000: $\frac{38754.85}{3000}=(1 +\frac{r}{4})^{80}$. So, $12.918283=(1 +\frac{r}{4})^{80}$.

Step3: Take the 80th root of both sides

$12.918283^{\frac{1}{80}}=1+\frac{r}{4}$. Using a calculator, $12.918283^{\frac{1}{80}}\approx1.03$. Then $1.03 = 1+\frac{r}{4}$.

Step4: Solve for $r$

Subtract 1 from both sides: $0.03=\frac{r}{4}$. Multiply both sides by 4 to get $r = 0.12$ or $12%$.

Answer:

$12$

2.

Explanation:

Step1: Recall the half - life formula

The formula for radioactive decay is $A=A_0(\frac{1}{2})^{\frac{t}{T}}$, where $A_0$ is the initial amount, $t$ is the time elapsed, and $T$ is the half - life. Here, $A_0 = 4$, $t = 1000$, and $T = 1620$.

Step2: Substitute the values into the formula

$A = 4(\frac{1}{2})^{\frac{1000}{1620}}$.

Step3: Calculate the value

First, calculate $\frac{1000}{1620}\approx0.6173$. Then $(\frac{1}{2})^{0.6173}\approx0.657$. Multiply by 4: $A\approx4\times0.657 = 2.628\approx2.6$.

Answer:

$2.6$

3.

Explanation:

Step1: Recall the formula for earthquake intensity

The magnitude $M$ of an earthquake is related to the intensity $I$ by the formula $M=\log(\frac{I}{I_0})$, where $I_0$ is a reference intensity. If $M_1$ and $M_2$ are the magnitudes of two earthquakes, then $M_1 - M_2=\log(\frac{I_1}{I_2})$. Here, $M_1 = 7.9$ and $M_2 = 4.1$. So, $7.9−4.1=\log(\frac{I_1}{I_2})$.

Step2: Simplify the left - hand side

$3.8=\log(\frac{I_1}{I_2})$.

Step3: Convert from logarithmic to exponential form

By the definition of logarithms, if $\log(x)=y$, then $x = 10^y$. So, $\frac{I_1}{I_2}=10^{3.8}\approx6310$.

Answer:

$6310$

4.

Explanation:

Step1: Rewrite 216 as a power of 6

We know that $216 = 6^3$. So the equation $6^{7x}=216$ becomes $6^{7x}=6^3$.

Step2: Set the exponents equal

Since the bases are the same, we can set the exponents equal: $7x=3$.

Step3: Solve for $x$

Divide both sides by 7: $x=\frac{3}{7}\approx0.43$.

Answer:

$\frac{3}{7}$

5.

Explanation:

Step1: Isolate the exponential term

Starting with $70 - e^{\frac{x}{2}}=65$, subtract 70 from both sides: $-e^{\frac{x}{2}}=65 - 70=-5$. Then multiply both sides by - 1 to get $e^{\frac{x}{2}}=5$.

Step2: Take the natural logarithm of both sides

$\ln(e^{\frac{x}{2}})=\ln(5)$.

Step3: Use the property of logarithms

Since $\ln(e^a)=a$, we have $\frac{x}{2}=\ln(5)$. Multiply both sides by 2: $x = 2\ln(5)\approx2\times1.6094 = 3.22$.

Answer:

$3.22$

6.

Explanation:

Step1: Simplify the logarithmic equation

Given $2\log_2x = 14$, divide both sides by 2: $\log_2x = 7$.

Step2: Convert from logarithmic to exponential form

By the definition of logarithms, if $\log_a b=c$, then $b=a^c$. So, $x = 2^7=128$.

Answer:

$128$