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Logarithmic & Exponential Functions (Cambridge (CIE) IGCSE Additional Maths): Flashcards

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FrontExponential Functions
Define an exponential function.

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Cards in this collection (38)

Define an exponential function.
An exponential function is one in which the variable is the power, of the form $y = a^{x}$ with $a > 0$ .
That is what separates it from something like $y = x^{2}$ , where the variable is the base instead.
Why does every graph of $y = a^{x}$ pass through $(0, 1)$ ?
Because $a^{0} = 1$ for every allowed value of $a$ .
So whatever base is chosen, substituting $x = 0$ gives $y = 1$ .
Fill in the gaps about the shape of $y = a^{x}$ :
If $a > 1$ the graph shows exponential $______$ , and if $0 < a < 1$ it shows exponential $______$ .
The completed sentence is:
If $a > 1$ the graph shows exponential growth, and if $0 < a < 1$ it shows exponential decay.
A growth curve climbs as $x$ increases, while a decay curve falls away towards the $x$ -axis.
True or False?
However negative $x$ becomes, the graph of $y = a^{x}$ never touches the $x$ -axis.
True.
The $x$ -axis is an asymptote, so the curve gets steadily closer to it without ever reaching it.
A positive number raised to any power stays positive, so $y$ can never actually be zero.
Define $e$ , the exponential constant.
$e$ is an irrational number, roughly $2.718$ , used as the base of the exponential function $y = e^{x}$ .
Like every exponential graph, $y = e^{x}$ passes through $(0, 1)$ and has the $x$ -axis as an asymptote.
What is special about the graph of $y = e^{x}$ ?
Its gradient at every point equals its own value at that point.
So where the curve has height 5, the gradient of the curve there is also 5.
How is the graph of $y = e^{- x}$ related to $y = e^{x}$ ?
It is a reflection in the $y$ -axis.
Replacing $x$ by $- x$ swaps the left and right halves of the curve, turning a rising graph into a falling one.
In the model $y = A e^{k x}$ with $k > 0$ , what does $A$ represent, and how would you write decay instead?
$A$ is the initial value, the value of $y$ when $x = 0$ , because $e^{0} = 1$ .
Decay is written with a negative power, $y = A e^{- k x}$ , so it is the sign in the power that separates the two.
How would you read $\log_{5} 125 = 3$ aloud, in a way that explains what it means?
As the power you raise 5 to, in order to get 125, is 3.
Reading logarithms this way turns the notation back into the statement $5^{3} = 125$ , since a logarithm is the inverse of raising to a power.
Why are logarithms needed at all?
They solve equations where the unknown is the power.
You can spot the answer to $2^{x} = 8$ by inspection, but $2^{x} = 10$ has no obvious solution, and it is exactly $\log_{2} 10$ .
Fill in the gaps about logarithm abbreviations:
$\log_{10} x$ is written as $\log x$ or as $______$ , and $\log_{e} x$ is written as $______$ .
The completed sentence is:
$\log_{10} x$ is written as $\log x$ or as $\lg x$ , and $\log_{e} x$ is written as $\ln x$ .
Bases 10 and $e$ are common enough to have earned their own shorthand.
True or False?
$(\log x)^{2}$ and $\log x^{2}$ mean the same thing.
False.
$(\log x)^{2}$ means take the logarithm and then square it, while $\log x^{2}$ means square $x$ first and then take the logarithm.
The brackets decide which operation happens first, so the two give different values.
Define $\ln$ .
$\ln$ is the natural logarithm, the logarithm with base $e$ , so $\ln x \equiv \log_{e} x$ .
It is a function, not a number, so it cannot be cancelled or multiplied out the way a coefficient can.
Fill in the three standard results:
$\ln 1 =______$
$\ln e =______$
$\ln e^{x} =______$
The completed results are:
$\ln 1 = 0$
$\ln e = 1$
$\ln e^{x} = x$
Each follows from the definition, since $\ln$ asks what power of $e$ gives the number.
For which values of $x$ is $\ln x$ defined?
Only for positive $x$ .
No power of $e$ can produce zero or a negative number, so $\ln 0$ and the logarithm of a negative number do not exist.
Solve $e^{2 x} = 5$ , leaving your answer as an exact value.
Take the natural logarithm of both sides and use $\ln e^{a} = a$ :
$2 x = \ln 5$
Dividing by 2 gives $x = \frac{\ln 5}{2}$ , which is left in this form because it is exact.
Complete the three laws of logarithms, for $a$ , $x$ , $y > 0$ :
$\log_{a} x y = \log_{a} x +______$
$\log_{a} \frac{x}{y} = \log_{a} x -______$
$\log_{a} x^{m} =______$
The completed laws are:
$\log_{a} x y = \log_{a} x + \log_{a} y$
$\log_{a} \frac{x}{y} = \log_{a} x - \log_{a} y$
$\log_{a} x^{m} = m \log_{a} x$
Multiplying becomes adding, dividing becomes subtracting, and a power comes down to the front.
Why do the laws of logarithms look like the laws of indices?
Because a logarithm is a power, so each log law is an index law written the other way round.
For instance $a^{x} \times a^{y} = a^{x + y}$ turns multiplying into adding powers, which is exactly what $\log_{a} x y = \log_{a} x + \log_{a} y$ says.
True or False?
$\log_{a} (x + y) = \log_{a} x + \log_{a} y$
False.
The addition law works on a product inside the logarithm, not a sum, so it is $\log_{a} x y$ that splits into $\log_{a} x + \log_{a} y$ .
A logarithm of a sum cannot be broken apart at all.
Complete these standard results, for any base $a$ :
$\log_{a} a =______$
$\log_{a} 1 =______$
$a^{\log_{a} x} =______$
The completed results are:
$\log_{a} a = 1$
$\log_{a} 1 = 0$
$a^{\log_{a} x} = x$
The last one says that raising to a power and taking a logarithm undo each other.
Write $2 \log 4 - \log 2$ as a single logarithm.
Bring the 2 up as a power first, then use the subtraction law:
$2 \log 4 - \log 2 = \log 16 - \log 2 = \log \frac{16}{2}$
So the expression is $\log 8$ .
After solving an equation involving $\log (x + k)$ , why must you check your solutions?
Because a logarithm is only defined for a positive argument, so any solution making $x + k$ zero or negative has to be rejected.
The algebra itself will not warn you: an invalid value can come out looking exactly like a valid one.
State the change of base formula, and say when it is needed.
The formula is $\log_{a} x = \frac{\log_{b} x}{\log_{b} a}$ , where $b$ is any base you choose.
It is needed when an equation contains logarithms with different bases, since the laws of logarithms only apply once all the bases match.
Find $\log_{8} 32$ without a calculator.
Both 8 and 32 are powers of 2, so change to base 2:
$\log_{8} 32 = \frac{\log_{2} 32}{\log_{2} 8} = \frac{5}{3}$
Choosing a base that both numbers are powers of turns the logarithms into whole numbers you already know.
What makes an equation an exponential equation?
The unknown appears as a power rather than as a base.
So $2^{x} = 8$ is exponential, whereas $x^{2} = 8$ is not.
Solve $5^{2 x} = 125$ without a calculator.
Write both sides as powers of the same base, since $125 = 5^{3}$ :
$5^{2 x} = 5^{3}$
Once the bases match the powers must be equal, so $2 x = 3$ and $x = \frac{3}{2}$ .
Complete the index rewrites that turn $4^{x} - 3 (2^{x + 1}) + 9 = 0$ into a quadratic:
$4^{x} = (______)^{2}$
$2^{x + 1} = 2 (______)$
Both gaps are $2^{x}$ :
$4^{x} = (2^{x})^{2}$
$2^{x + 1} = 2 (2^{x})$
The first uses $4 = 2^{2}$ and the second splits the power as $2^{x} \times 2^{1}$ , leaving a quadratic in $2^{x}$ .
How do you solve $2^{x} = 3$ ?
Take logarithms of both sides, then bring the power down to the front:
$x \ln 2 = \ln 3$
Dividing gives $x = \frac{\ln 3}{\ln 2}$ , which is about $1.58$ .
A question asks for an exact answer. Why give $x = \frac{\ln 3}{\ln 2}$ rather than $1.58$ ?
Because $1.58$ is a rounded value, so it is not equal to the solution, only close to it.
Leaving the answer in logarithm form keeps it exact, and nothing is lost by not evaluating it.
True or False?
Every exponential equation can be solved by writing both sides as powers of the same base.
False.
It works when the numbers are related, as with $5^{2 x} = 125$ , because $125$ is a power of 5.
But 3 is not a power of 2, so $2^{x} = 3$ cannot be done that way and needs logarithms instead.
How do you turn $y = a x^{n}$ into a linear relationship?
Take logarithms of both sides, split the product, and bring the power down:
$\ln y = \ln a + n \ln x$
That is a straight line once you treat $\ln y$ and $\ln x$ as the two variables.
For $y = a x^{n}$ linearised as $\ln y = \ln a + n \ln x$ , match it to $Y = m X + c$ :
$Y =______$
$X =______$
$m =______$
$c =______$
The completed matching is:
$Y = \ln y$
$X = \ln x$
$m = n$
$c = \ln a$
Both variables are logged here, which is what makes a power law come out straight.
The graph of $\ln y$ against $\ln x$ has gradient $- 1.2$ and intercept $0.3$ . What are $n$ and $a$ in $y = a x^{n}$ ?
The gradient is $n$ directly, so $n = - 1.2$ .
The intercept is $\ln a$ , so solve $\ln a = 0.3$ by raising $e$ to each side, giving $a = e^{0.3}$ .
Why is $a = e^{0.3}$ left in that form rather than worked out?
Because it is the exact value, and the question may require exact form.
Evaluating it would round the answer, which loses accuracy for no gain unless a decimal is actually asked for.
How do you turn $y = A b^{x}$ into a linear relationship?
Take logarithms of both sides, split the product, and bring the power down:
$\lg y = \lg A + x \lg b$
Here the power was already $x$ , so it comes down as a plain $x$ rather than as a logarithm.
For $y = A b^{x}$ linearised as $\lg y = \lg A + x \lg b$ , match it to $Y = m X + c$ :
$Y =______$
$X =______$
$m =______$
$c =______$
The completed matching is:
$Y = \lg y$
$X = x$
$m = \lg b$
$c = \lg A$
Note that $X$ is $x$ itself, not its logarithm, and that both constants sit inside logarithms.
True or False?
To linearise either $y = a x^{n}$ or $y = A b^{x}$ , you plot $\log y$ against $\log x$ .
False.
For the power law $y = a x^{n}$ you do plot $\log y$ against $\log x$ , logging both variables.
But for the exponential law $y = A b^{x}$ you plot $\log y$ against $x$ itself, because the unknown is already in the power.
A graph of $\lg y$ against $x$ gives $\lg b = 1$ and $\lg A = 3$ . Find $b$ and $A$ .
Undo each logarithm by raising 10 to both sides, since $\lg$ has base 10:
$b = 10^{1} = 10$
$A = 10^{3} = 1000$
Had the working used $\ln$ instead, you would raise $e$ to each side in the same way.

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