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\(a=\log_{10}x\) 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.

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 \(x<2\)

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

Solve the equation for \(x\)

\(\log_{10}x = -3\)

Solution.

\(x = 10^{-3} \) \(=0.001 \) <---- Answer

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

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

Solve the equation for \(x\)

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

Solution.

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

\(4x=10\), thus \(x=2.5 \) <---- Answer

\(y=\log_{10}x\)

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 \(x>0\).

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 __

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 \(t=4\) ... \(Q=900\)

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

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 \( 11.4 \%\).

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

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

\(Q=Q_0\times\left(\frac{1}{2}\right)^{t/T_{\text{half}}} =300 \times \left(\frac{1}{2}\right)^{t/8}\)

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