Logarithmic Functions

Simplify the expression ${10}^{{\log }_{100}\left(x\right)}$.

Answer:

Step 1: If that were a logarithm base 10 then the answer would be $x$ using properties of inverse functions. So the idea is to use the Proportion Rule (also known as the change of base formula) to make it into a base 10 logarithm first. There are two ways to think about doing this, and you get the same answer either way.

The first way to think about it is to use the fact that 10 is the number you are raising to a power to get:

$$\begin{gathered} {\log }_{10}\left(x\right)=\frac{{\log }_{100}\left(x\right)}{{\log }_{100}\left(10\right)} \\ =\frac{{\log }_{100}\left(x\right)}{{\log }_{100}\left({100}^{\frac{1}{2}}\right)} \\ =\frac{{\log }_{100}\left(x\right)}{\frac{1}{2}{\log }_{100}\left(100\right)} \\ =2·{\log }_{100}\left(x\right) \end{gathered}$$

The second way is to look at the logarithm and see that it is base 100, and use that to get:

and then solve for ${\log }_{10}\left(x\right)$ to get that ${\log }_{10}\left(x\right)=2·{\log }_{100}\left(x\right).$ Both methods work, and you can use the one that is easiest for you to understand and remember.

So you have

${\log }_{100}\left(x\right)=\frac{1}{2}{\log }_{10}\left(x\right)$

Step 2: Now using properties of exponents,

$$\begin{gathered} {10}^{{\log }_{100}\left(x\right)}={10}^{\frac{1}{2}{\log }_{10}\left(x\right)} \\ ={\left({10}^{{\log }_{10}\left(x\right)}\right)}^{\frac{1}{2}} \\ =x^{\frac{1}{2}} \end{gathered}$$

So the expression ${10}^{{\log }_{100}\left(x\right)}$ completely simplified is

List at least 3 points on the graph of $f\left(x\right)={\log }_3\left(x\right)$ without graphing the function or using a calculator.

Answer:

This problem looks trickier than it actually is. You already know that the inverse of $f\left(x\right)$ is $f^{-1}\left(x\right)=3^x$, and that if $(a,b)$ is a point on the graph of $f\left(x\right)$ then $(b,a)$ is a point on the graph of $f^{-1}\left(x\right)$.

Step 1: Exponential functions have a y intercept at $\left(0,1\right)$, so the point $(1,0)$ is on the graph of $f\left(x\right)$.

Step 2: To get two more points on the graph, evaluate points on the graph of $f^{-1}\left(x\right)$. Choosing two random values, $f^{-1}\left(2\right)=3^2=9$.

Step 3: Evaluate another point on the graph of $f^{-1}\left(x\right)=3^x.$ $f^{-1}\left(3\right)=3^3=27.$

Step 4: So the points $(2,9)$ and $\left(3,27\right)$ are on the graph of $f^{-1}\left(x\right)$. So you know that the points $(9,2)$ and $\left(27,3\right)$ are points on the graph of $f\left(x\right)$.

So without using a calculator, you can see that three points on the graph of $f\left(x\right)$ are $\left(1,0\right)$, $\left(9,2\right)$, and $\left(27,3\right)$.

What if instead, the base was a fractional power of 2? For example if $f\left(x\right)={\log }_2\left(x\right)$ but now you have

$m\left(x\right)={\log }_{\frac{1}{2}}\left(x\right)$, $n\left(x\right)={\log }_{\frac{1}{4}}\left(x\right)$, and $p\left(x\right)={\log }_{\frac{1}{8}}\left(x\right)$, then

which means that again they are constant multiples of $f\left(x\right)$, but they should be flipped over the x axis as well.

So as you can see, these three new functions are

• decreasing,
• concave up,
• have $x=0$ as the vertical asymptote, and
• all have the same x intercept at $\left(1,0\right)$.

Sounds are measured on a logarithmic scale using the unit, decibels (dB). Sound can be modeled using the equation:

Where

• $p$ is the power of the sound,
• $p_0$ is the smallest sound a person can hear, and
• $d\left(p\right)$ is the number of decibels for the power $p$:

Say you are thinking of buying a new speaker. Speaker A says it has a noise rating of 50 decibels, while speaker B says it has a noise rating of 75 decibels. How much more intense is the sound from speaker B than from speaker A?

Answer:

Step 1: For comparison, call $d_A\left(p\right)$ the decibel level of speaker A, and $d_B\left(p\right)$ the decibel level of speaker B. From the information given, you know that:

• speaker A has $50=10\log \left(\frac{A}{p_0}\right)$ where $A$ is the power of speaker A
• speaker B has $75=10\log \left(\frac{B}{p_0}\right)$ where $B$ is the power of speaker B

Step 2: Taking the equation for speaker A and writing it in terms of $p_0$ will let you substitute it into the equation for speaker B. So

$$\begin{gathered} 50=10\log \left(\frac{A}{p_0}\right) \\ 5=\log \left(\frac{A}{p_0}\right) \\ {10}^5={10}^{\log \left(\frac{A}{p_0}\right)} \\ {10}^5=\frac{A}{p_0} \\ {p}_0=\frac{A}{{10}^5}. \end{gathered}$$

Step 3: Substituting this into the equation for speaker B,

$$\begin{gathered} 75=10\log \left(\frac{B}{p_0}\right) \\ 7.5=\log \left(\frac{B}{\frac{A}{{10}^5}}\right) \\ 7.5=\log \left(\frac{{10}^{5}B}{A}\right) \\ {10}^{7.5}={10}^{\log \left(\frac{{10}^{5}B}{A}\right)} \\ {10}^{7.5}=\frac{{10}^{5}B}{A} \\ B=\frac{{10}^{7.5}A}{{10}^5} \\ B\approx 316A \end{gathered}$$

So, the sound from speaker B is about 316 times more intense than that of speaker A!

Earthquakes are measured on a logarithmic scale called the Richter scale. The magnitude of an earthquake is a measure of how much energy is released. Here $A_0$ is amplitude of the smallest wave that a seismograph (the device that measures how much the earth is moving) can measure. Then the formula for the Richter scale measurement of an earthquake is

where $A$ measures the amplitude of the earthquake wave. In general an earthquake measures between 2 and 10 on the Richter scale. Ones that scores less than 5 on the scale are considered relatively minor, and anything above an 8 on the scale is likely to cause quite a bit of damage. In fact, a magnitude 5 earthquake is 10 times as powerful as a magnitude 4 earthquake.

Suppose that an earthquake in Indiana had a magnitude of 8.1 on the Richter scale, but one on the same day in California was 1.26 times as intense. What was the magnitude of the earthquake in California?

Answer:

Step 1: Using the definition of the Richter scale, and using $A_{Ind}$ for the amplitude of the Indiana earthquake, the earthquake in Indiana had

$$\begin{gathered} {10}^{8.1}={10}^{\log \left(\frac{A_{Ind}}{A_0}\right)} \\ {10}^{8.1}=\frac{A_{Ind}}{A_0} \\ {A}_{Ind}={10}^{8.1}{A}_0. \end{gathered}$$

Now you can use the fact that the California earthquake was 1.26 times as intense as the Indiana one, or in other words, if $A_{Cal}$ is the amplitude of the California earthquake, then $A_{Cal}=1.26A_{Ind}$. So

$$\begin{gathered} R\left(A_{Cal}\right)=\log \left(\frac{A_{Cal}}{A_0}\right) \\ =\log \left(\frac{1.26A_{Ind}}{A_0}\right) \\ =\log \left(\frac{1.26·{10}^{8.1}{A}_0}{A_0}\right) \\ =\log \left(1.26·{10}^{8.1}\right)\approx 8.2 \end{gathered}$$

That means the earthquake in California measured about 8.2 on the Richter scale.