# Section 9C

### exercises

...

Review of logarithms for Exercises 11-26

$$a=\log_{10}x$$ means nothing but $$10^a=x$$

Example. $$2=\log_{10}100$$ because $$10^2=100$$

Useful rules to apply

$$\log_{10}10^x=x$$

$$10^{\log_{10}x} =x \ \ \ (x>0)$$

$$\log_{10}(xy)=\log_{10}x+\log_{10}y \ \ \ (x>0 \ \text{and} \ y>0)$$

$$\log_{10}a^x=x\log_{10}a \ \ \ (a>0)$$

More Details in Brief Review on p488 and p542 of the textbook.

Logarithms. Exercise 12

Solve the equation for $$x$$

$$10^x=23$$

Solution.

$$x = \log_{10}23$$ $$\approx 1.36$$ <---- Answer

Check that roughly:
since $$10^1=10$$, we know $$x>1$$
since $$10^2=100$$, we know $$x<2$$

Logarithms. Exercise 16

Solve the equation for $$x$$

$$3\times 4^x=180$$

Solution.

$$4^x=180/3$$

$$x\times \log_{10}4=\log_{10}(180/3)$$

$$x=\frac{\log_{10}(180/3)}{\log_{10}4}$$ $$\approx 2.95$$ <---- Answer

Logarithms. Exercise 20

Solve the equation for $$x$$

$$\log_{10}x = -3$$

Solution.

$$x = 10^{-3}$$ $$=0.001$$ <---- Answer

Indeed, $$\log_{10}10^a = a$$, thus $$\log_{10}10^{-3} = -3$$

Logarithms. Exercise 24

Solve the equation for $$x$$

$$\log_{10}(3x) = 5.1$$

Solution.

$$3x=10^{5.1}$$

$$x= \frac{10^{5.1}}{3}$$ $$\approx 41964.18$$ <---- Answer

Logarithms. Exercise 26

Solve the equation for $$x$$

$$4\log_{10}(4x) = 4$$

Solution.

$$\log_{10}(4x) = 1$$

$$4x=10$$, thus $$x=2.5$$ <---- Answer

Remark on logarithms

$$y=\log_{10}x$$

Is another example of a function .

To a value of the independent variable $$x$$, it assigns the value of dependent variable $$y$$ such that

$$10^y = x$$

Not every $$x$$ is possible though. The domain of this function is $$x>0$$.

Exponential growth and decay. Exercise 28

The number of restaurants in a city that had 800 restaurants in 2013 increases at a rate of 3% per year.

a. Create the exponential function to model the situation described.

Solution. We want
$$Q=Q_0\times (1+r)^t$$

with $$r=3\%=0.03$$, and $$Q_0=800$$ we obtain

$$Q=800 \times 1.03^t,$$

where $$t$$ is time after 2013 in years

Exercise 28 ... continued

The number of restaurants in a city that had 800 restaurants in 2013 increases at a rate of 3% per year.

b. Create a table showing the values of the quantity $$Q$$ for the first units of time, years in our case

Solution. We calculate using our formula $$Q=800 \times 1.03^t$$

Year 2013 $$t=0$$ ... $$Q=800$$

Year 2014 $$t=1$$ ... $$Q=824$$

Year 2015 $$t=2$$ ... $$Q=849$$

Year 2016 $$t=3$$ ... $$Q=874$$

Year 2017 $$t=4$$ ... $$Q=900$$

Exercise 28 ... continued

The number of restaurants in a city that had 800 restaurants in 2013 increases at a rate of 3% per year.

c. Make a graph of the exponential function

Exercise 36

If the price of gold decreases at a monthly rate of $$1\%$$, by what percentage does it decrease in a year?

Solution. The price of gold is described by
$$Q=Q_0\times (1-0.01)^t = Q_0 \times 0.99^t$$,

where $$t$$ is the time in months.

After a year ($$t=12$$) the price changes by the factor of

$$0.99^{12} \approx 0.886 = 1-0.114$$.

We thus see a drop of the price by $$11.4 \%$$.

Exercise 42

Assume that aspirin has a half-life of 8 hours in the bloodstream. At 12:00 noon you take a 300-milligram dose of aspirin. a. How much aspirin will be in your bloodstream at 6pm the same day?

Solution. The amount of aspirin in your bloodstream is described by the formula
$$Q=Q_0\times\left(\frac{1}{2}\right)^{t/T_{\text{half}}} =300 \times \left(\frac{1}{2}\right)^{t/8}$$

where $$Q$$ is the amount in milligrams, and $$t$$ is the time in hours since 12:00 noon

At 6pm ($$t=6$$ ) there is

$$300 \times \left(\frac{1}{2}\right)^{6/8}$$ $$\approx 178$$ milligram <--- Answer

Exercise 42 ... continued

Assume that aspirin has a half-life of 8 hours in the bloodstream. At 12:00 noon you take a 300-milligram dose of aspirin. b. Estimate when the amount of aspirin will decay to $$5\%$$ of its original amount.

Solution. The amount of aspirin in your bloodstream is described by the formula
$$Q=Q_0\times\left(\frac{1}{2}\right)^{t/T_{\text{half}}} =300 \times \left(\frac{1}{2}\right)^{t/8}$$

We solve this equation for $$t$$:

$$\log_{10}(Q/300) = (t/8) \times \log_{10} \left(\frac{1}{2}\right)$$

$$t=8 \times \frac{\log_{10}(Q/300)}{\log_{10} \left(\frac{1}{2}\right)}$$

Exercise 42 ... continued

Assume that aspirin has a half-life of 8 hours in the bloodstream. At 12:00 noon you take a 300-milligram dose of aspirin. b. Estimate when the amount of aspirin will decay to $$5\%$$ of its original amount.

Solution. So far we have
$$t=8 \times \frac{\log_{10}(Q/300)}{\log_{10} \left(\frac{1}{2}\right)}$$

We want $$Q$$ to be $$5\%$$ of $$300$$, thus

we search for $$t$$ such that $$Q/300 = 0.05$$

That is $$t=8 \times \frac{\log_{10}(0.05)}{\left(\log_{10}\frac{1}{2}\right)} \approx 34.6$$ Answer: that happens 34.6 hours later, around 10:30pm next day