Home | | Latest | About | Random
# Logarithms.
Let us review the definition of logarithm:
The logarithm function of base $b$, where $b > 0$ and $b \neq 1$, is denote by $y =\log_{b}(x)$ and is defined by $$
y=\log_{b}x \quad\text{if and only if}\quad x=b^{y}
$$
that is to say, the logarithm of base b of the number $x$, is the power $y$ we need such that $b^{y}$ gives $x$. The domain of $y=\log_{b}(x)$ is for $x > 0$.
We will use the special notation $\ln$ to denote $\log_{e}$ which is the natural log, with base $e\approx 2.718281828459....$ In particular we have $$
y=e^{x} \quad\text{if and only if} \quad x=\ln (y)
$$Some logarithm rules:
(1) $\log_{b}(xy)=\log_{b}(x) + \log_{b}(y)$
(2) $\log_{b}(x^m) = m\log_{b}(x)$
(3) $\displaystyle\log_{b}(x) = \frac{\log_{c}(x)}{\log_{c}(b)}$
![[---images/---assets/---icons/question-icon.svg]] Find the exact value of each logarithm without a calculator.
(1) $\log_{2}0$
(2) $\log_{5}125$
(3) $\log_{3} \frac{1}{9}$
(4) $\log_{\frac{1}{3}} 9$
(5) $\log_{\sqrt{2}}4$
(6) $\log_{5} \sqrt[3]{25}$
(7) $\ln \sqrt{e}$
(8) $\ln e^{3}$
![[---images/---assets/---icons/question-icon.svg]] Solve for $x$ in each of the following.
(1) $\log_{3}x = 2$
(2) $\log_{3}(3x-2)=2$
(3) $\log_{x}(\frac{1}{8})=3$
(4) $e^{-2x + 1}=13$
(5) $\log_{5}(x^2 + x + 4) = 2$
(6) $\log_{2} 8^{x} = -3$
![[---images/---assets/---icons/question-icon.svg]] Solve for $y$ in terms of $x$ in the following $\ln x = \ln(y+1)+\ln(y+2)$.
![[---images/---assets/---icons/question-icon.svg]] Find the exact value of $\log_{2}3\cdot \log_3 4 \cdot \log_{4}\cdot 5\cdot\log_{5} 6 \cdot\log_{6} 7\cdot\log_{7}8$ without a calculator.
## Example. Blood alcohol content (BAC) and driving.
**Blood alcohol content** (BAC) is a measure of the amount of alcohol in a person's blood stream. A BAC of 0.02% means a person has 2 parts of alcohol for 10000 parts blood in their body.
**Relative risk** is defined to be the likelihood of one event happening divided by the likelihood of another event happening. So for example, if a person with BAC of 0.02% is 1.4 times as likely to get into an accident while driving than a person who has not been drinking, we say the person with BAC of 0.02% has a relative risk of 1.4.
Many medical research suggests that the relative risk factor $R$ of having an accident while driving a car can be modeled by an equation of the form $$
R = A e^{kx}
$$
where $x$ is the percent of concentration of alcohol in the bloodstream and $k$ is some constant.
(One can see such a study like this here: [Voas, et al.](https://www.researchgate.net/publication/223136111_Alcohol-Related_Risk_of_Driver_Fatalities_An_Update_Using_2007_Data))
(a) Using the definition of relative risk $R$, what is the constant $A$ when the percent concentration of alcohol in the blood stream is 0 (i.e. when a person has not been drinking)?
(b) Some research suggests that the relative risk of a person having an accident with a BAC of 0.02% is 1.4. Find the constant $k$.
(c) Using this value of $k$ you found, what is relative risk $R$ when the concentration is 0.17%?
(d) Using this value of $k$, what BAC corresponds to a relative risk $R$ of 100?
(e) If the law suggests that having a relative risk $R$ of 4 or more, one should not have driving privileges, at what BAC would a driver be arrested and charged with a DUI (driving under the influence)?
(f) Look up at what BAC would one receive a DUI in California.
Random note. Above a BAC of 0.3% is likely to put one into a coma and possibly death. [Website: Government of South Australia](https://www.sahealth.sa.gov.au/wps/wcm/connect/public+content/sa+health+internet/conditions/alcohol/blood+alcohol+concentration+bac+and+the+effects+of+alcohol)