International A Level (IAL)MathsEdexcelPure 3FlashcardsExponentials & LogarithmsExponential & Logarithms

Exponential & Logarithms (Edexcel International A Level (IAL) Maths: Pure 3): Flashcards

Exam code: YMA01

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Fronte & ln
Define e, Euler's number.

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Cards in this collection (13)

Define e, Euler's number.
$e$ is an irrational number, roughly $2.718$ , which serves as the base of natural logarithms.
Like $π$ , it cannot be written exactly as a fraction or as a terminating decimal.
How is the graph of $y = e^{- x}$ related to $y = e^{x}$ ?
It is the reflection of $y = e^{x}$ in the $y$ -axis.
The pair are in the form $y = f (x)$ and $y = f (- x)$ , which is what produces that reflection.
What does $\ln x$ mean?
It is the natural logarithm, the logarithm to base $e$ : $\ln x \equiv \log_{e} x$ .
Everything that is true of logarithms in general is therefore true of $\ln$ as well.
True or False?
$\ln$ is a number, in the way that $e$ and $π$ are.
False.
$\ln$ is a function, so it needs something to act on before it means anything at all.
$\ln 5$ is a number, but $\ln$ by itself is not, any more than $\sin$ by itself is.
Complete these three results:
$\ln 1 =______$
$\ln e =______$
$\ln e^{x} =______$
The completed results are:
$\ln 1 = 0$
$\ln e = 1$
$\ln e^{x} = x$
Each one comes straight from the meaning of a logarithm applied to base $e$ .
What is the relationship between $e^{x}$ and $\ln x$ ?
They are inverse functions, so each one undoes the other.
That is why $e^{\ln x} = x$ , and why applying one of them to both sides of an equation strips the other away.
An equation reads $e^{f (x)} = g (x)$ . How do you get at $f (x)$ ?
Take $\ln$ of both sides, which leaves $f (x) = \ln g (x)$ .
It works the other way round too: $\ln f (x) = g (x)$ becomes $f (x) = e^{g (x)}$ .
True or False?
$\ln x$ has no value when $x$ is zero or negative.
True.
No power of $e$ produces zero or a negative number, so there is nothing for the logarithm to return.
If $y = e^{k x}$ then $\frac{d y}{d x} =______$ , and if $y = e^{- k x}$ then $\frac{d y}{d x} =______$ .
If $y = e^{k x}$ then $\frac{d y}{d x} = k e^{k x}$ , and if $y = e^{- k x}$ then $\frac{d y}{d x} = - k e^{- k x}$ .
The constant from the power comes down as a multiplier, and $y = e^{x}$ is simply the case $k = 1$ .
What stays the same when you differentiate $e^{k x}$ ?
The exponential part itself: $e^{k x}$ reappears in the derivative unchanged.
Only a constant multiplier is added in front, which is what makes exponential derivatives unusually simple.
What is the derivative of $y = e^{- 3 x}$ ?
$\frac{d y}{d x} = - 3 e^{- 3 x}$ .
The minus sign comes down with the $3$ , so a decay curve has a negative gradient everywhere along it.
True or False?
The gradient of $y = e^{x}$ is never zero.
True.
The gradient equals $e^{x}$ , which is positive for every value of $x$ .
So the curve is always increasing and has no stationary points at all.
How do you find the gradient of $y = e^{2 x}$ at $x = 0$ ?
Differentiate to get $\frac{d y}{d x} = 2 e^{2 x}$ , then substitute $x = 0$ .
Since $e^{0} = 1$ , the gradient there is $2$ .

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