# Displacement, Velocity and Distance Travelled by Natural Logarithmic Equations

A particle is moving in a straight line, starting from the origin. At time t seconds the particle has a displacement of x metres from the origin and a velocity v. The displacement is described as a logarithmic expression in terms of time t. Find an expression for v, the initial velocity, when the particle comes to rest and the distance travelled by the particle in the first three seconds.

# Exponential Growth and Decay using Logarithms

It has been known that how exponential functions can be used to model a variety of growth and decay situations. These included the growth of populations and the decay of radioactive substances. In this lesson we consider more growth and decay problems, focusing particularly on how logarithms can be used in there solution. Population Growth […]

# Graphing Natural Logarithmic Functions

The inverse function of $y=e^x$ is $y=\log_{e}{x}$. Therefore $y=\log_{e}{x}$ is an inverse function, it is a reflection of $y=e^x$ in the line $y=x$. The graphs of $y=e^x$ is $y=\log_{e}{x}$: \begin{array}{|c|c|c|} \require{color} \hline & y=e^x & \color{red}y =\log_{e}{x} \\ \hline \text{domain} & x \in \mathbb{R} & \color{red}x \gt 0 \\ \hline \text{range} & y \gt 0 […]

# Graphing Logarithmic Functions

The inverse function of $y=a^x$ is $y=\log_{a}{x}$. Therefore $y=\log_{a}{x}$ is an inverse function, it is a reflection of $y=a^x$ in the line $y=x$. The graphs of $y=a^x$ is $y=\log_{a}{x}$ for $0 \lt a \lt 1$: The graphs of $y=a^x$ is $y=\log_{a}{x}$ for $a \gt 1$: \begin{array}{|c|c|c|} \require{color} \hline & y=a^x & \color{red}y =\log_{a}{x} \\ \hline […]

# Logarithm Change of Base Rule

$$\log_{b}{a} = \dfrac{\log_{c}{a}}{\log_{c}{b}}$$ $$\text{for }a,b,c>0 \text{ and } b,c \ne 1$$ For example, \begin{align} \log_{3}{8} &= \dfrac{\log_{2}{8}}{\log_{2}{3}} \\ &= \dfrac{\log_{5}{8}}{\log_{5}{3}} \\ &= \dfrac{\log_{10}{8}}{\log_{10}{3}} \\ &\vdots \\ &= 1.8927 \cdots \\ \end{align} $\textit{Proof:}$ \( \begin{align} \displaystyle \text{Let } \log_{b}{a} &= x \cdots (1)\\ b^x &= a \\ \log_{c}{b^x} &= \log_{c}{a} &\text{taking logarithm in base […]

# Exponential Inequalities using Logarithms

Inequalities are worked in exactly the same way except that there is a change of sign when dividing or multiplying both sides of the inequality by a negative number. \begin{array}{|c|c|c|} \hline \log_{2}{3}=1.6>0 & \log_{5}{3}=0.7>0 & \log_{10}{3}=0.5>0 \\ \hline \log_{2}{2}=1>0 & \log_{5}{2}=0.4>0 & \log_{10}{2}=0.3>0 \\ \hline \log_{2}{1}=0 & \log_{5}{1}=0 & \log_{10}{1}=0 \\ \hline \log_{2}{0.5}=-1<0 & \log_{5}{0.5}=-0.4<0 […]

# Exponential Equations using Logarithms

We can find solutions to simple exponential equations where we could make equal bases and then equate exponents (indices). For example, $2^{x}=8$ can be written as $2^x = 2^3$. Therefore the solution is $x=3$. However, it is not always easy to make the bases the same such as $2^x=5$. In these situations, we use $\textit{logarithms}$ […]

# Logarithmic Equations

We can use the laws of logarithms to write equations in a different form. This can be particularly useful if an unknown appears as an index (exponent). $$2^x=7$$ For the logarithmic function, for every value of $y$, there is only one corresponding value of $x$. $$y=5^x$$ We can, therefore, take the logarithm of both sides […]