NSW Y12 Maths - Advanced Exponential and Log Function Exponential Functions

Resources for Exponential Functions

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Exponential Functions Theory

This subtopic is a revision of the subtopics in the year ll course. \\  The solutions of exponential equations, the equations of tangents to exponential functions and transformations of exponential functions will be covered.\\  \begin{multicols}{2}  \textbf{Example 1}\\ %24073 Solve \(2{e^x} = {e^{2x}}\)\\  \textbf{Example 1 solution}\\ $\begin{array}{c} 2 e^{x}=e^{2 x} \\ \operatorname{let} y=e^{x} \\ 2 y=y^{2} \\ y^{2}-2 y=0 \\ y(y-2)=0 \\ y=0, y=2 \\ e^{x}=0 \quad(\text { no solution }) \\ e^{x}=2 \rightarrow x=\ln 2 \end{array}$\\  \columnbreak \textbf{Example 2} %24700 \begin{itemize} \item[\bf{i)}]What is the \(y\) - intercept of the point \(P\) on the curve \(y = {e^{2x}} + 1\) \item[\bf{ii)}]Find \(\dfrac{{dy}}{{dx}}\) for this curve where \(x = 1\) \item[\bf{iii)}]Find the equation of the tangent to this curve \end{itemize}  \textbf{Example 2 solution}\\ $\begin{aligned} \textbf{i)}\quad &y=e^{2 x}+1\\ &\text{At}\ x=0,\ y=e^{0}+1 \rightarrow\left(0, 2\right)\\ \textbf{ii)}\quad &\frac{d y}{d x}=2 e^{2 x}+1\\ &\text{At}\ x=1, \frac{d y}{d x}=2 e^{2}+1\\ \textbf{iii)}\quad &y-\left(e^{2}+1\right)=\left(2 e^{2}+1\right)(x-1) \\ &y-e^{2}-1=\left(2 e^{2}+1\right) x-2 e^{2}-1 \\ &y=\left(2 e^{2}+1\right) x-e^{2} \end{aligned}$\\  \end{multicols}

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  • Exponential Functions - Video - Graphing Exponential Functions: Another Example

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