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Logarithms, Indices & Exponentials (Edexcel IGCSE Further Pure Maths): Flashcards

Exam code: 4PM1

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FrontExponential Functions
For the exponential function $f (x) = a^{x}$ , where $a > 0$ and $a \neq 1$ , what are the domain and the range?

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For the exponential function $f (x) = a^{x}$ , where $a > 0$ and $a \neq 1$ , what are the domain and the range?
The domain is the set of all real numbers: $x$ may take any value, positive, negative or zero.
The range is the set of all positive real numbers.
Complete the two points that every graph of $y = a^{x}$ passes through:
$(0,______) and (1,______)$
The completed points are:
$(0, 1) and (1, a)$
These come straight from $a^{0} = 1$ and $a^{1} = a$ , which hold whatever the value of the base.
Why does the graph of $y = a^{x}$ never cross the $x$ -axis?
Because a positive number raised to any power is still positive, so $a^{x}$ can get very close to zero but never reaches it.
The $x$ -axis is therefore a horizontal asymptote, and the curve has no maximum or minimum point.
True or False?
The graph of $y = a^{x}$ passes through $(0, 1)$ even when $0 < a < 1$ .
True.
$a^{0} = 1$ for every positive base, so a decreasing exponential meets the $y$ -axis at exactly the same point as an increasing one.
How does the shape of $y = a^{x}$ change when $a$ is between $0$ and $1$ ?
The curve decreases from left to right instead of increasing.
Everything else is unchanged: it stays above the $x$ -axis, keeps the $x$ -axis as its asymptote, and still has no maximum or minimum point.
Define e, the exponential constant.
$e$ is the irrational constant $2.718281 \dots$ , and $f (x) = e^{x}$ is known as the exponential function.
Because $e$ is irrational its decimal expansion never ends or repeats, so exact answers are left in terms of $e$ rather than rounded.
What is special about the gradient of the curve $y = e^{x}$ ?
At every point the gradient is equal to the $y$ value at that point, so $\frac{d y}{d x} = e^{x}$ .
No other function of the form $a^{x}$ behaves this way, and it is the reason $e$ is chosen as the base.
How is the graph of $y = e^{- x}$ related to the graph of $y = e^{x}$ ?
$y = e^{- x}$ is the reflection of $y = e^{x}$ in the $y$ -axis.
By the laws of indices $e^{- x} = {(\frac{1}{e})}^{x}$ , so it is a decreasing exponential whose base $\frac{1}{e}$ lies between $0$ and $1$ .
Complete the definition of a logarithm by filling in the missing index and result:
$\log_{a} b = x means a^{______} =______$
The completed definition is:
$\log_{a} b = x means a^{x} = b$
The base $a$ stays as the base on both sides, and the logarithm itself is always the index.
What does $\log_{5} 125 = 3$ tell you, put as a sentence about powers?
$\log_{5} 125 = 3$ says that $3$ is the power you raise $5$ to in order to get $125$ .
Reading a logarithm this way, as the power that produces the number, turns any log statement into an index statement you can check: $5^{3} = 125$ .
What do $\ln x$ and $\log x$ mean?
$\ln x$ is the natural logarithm $\log_{e} x$ , whose base is the constant $e$ .
$\log x$ written with no base means $\log_{10} x$ , because logarithms to base $10$ are used so often that the base is left off.
Why can you not take the logarithm of $0$ or of a negative number?
Because $\log_{a} x$ asks which power of $a$ gives $x$ , and a positive base raised to any power is always positive.
No index will ever produce $0$ or a negative result, so the domain of $\log_{a} x$ is $x > 0$ .
Define logarithmic function.
A logarithmic function has the form $f (x) = \log_{a} x$ , where on this course the base $a$ is always an integer greater than one.
Its range is the set of all real numbers, so the value of a logarithm can be positive, negative or zero.
Where does the graph of $y = \log_{a} x$ cross the axes, and where is its asymptote?
The graph crosses the $x$ -axis once, at $(1, 0)$ , and has no intercept on the $y$ -axis.
The $y$ -axis itself is a vertical asymptote, $x = 0$ , and the curve also passes through $(a, 1)$ .
True or False?
The graph of $y = \log_{a} x$ has a minimum point where it meets the $x$ -axis.
False.
The curve keeps falling to the left of that point, running steeply downwards alongside the vertical asymptote at the $y$ -axis.
A logarithmic graph has no maximum and no minimum point anywhere.
How are the graphs of $y = a^{x}$ and $y = \log_{a} x$ related?
Each is the reflection of the other in the line $y = x$ , because $a^{x}$ and $\log_{a} x$ are inverse functions.
Reflecting in $y = x$ swaps every $x$ coordinate with its $y$ coordinate, which is why an asymptote running along one axis becomes an asymptote running along the other.
What do $\log_{a} (a^{x})$ and $a^{\log_{a} x}$ both simplify to?
Both simplify to $x$ .
A logarithm asks which power of $a$ produces a number, so applying it to $a^{x}$ hands back the index $x$ ; going the other way, raising $a$ to that index rebuilds the original number.
Define base and index in the expression $a^{n}$ .
In $a^{n}$ the base is $a$ , the number being multiplied by itself, and the index is $n$ , which says how many times.
The index is also called the power or the exponent.
Complete the two index laws for multiplying and dividing, by filling in the missing indices:
$a^{m} \times a^{n} = a^{______} and a^{m} \div a^{n} = a^{______}$
The completed laws are:
$a^{m} \times a^{n} = a^{m + n} and a^{m} \div a^{n} = a^{m - n}$
Multiplying adds the indices and dividing subtracts them.
What must be true of two powers before you can add or subtract their indices?
The two powers must have the same base.
Those laws work by counting how many copies of that one base are being multiplied together, so an expression such as $2^{3} \times 3^{4}$ cannot be combined this way at all.
Complete these two index laws by filling in the missing indices:
${(a^{m})}^{n} = a^{______} and a^{- m} = \frac{1}{a^{______}}$
The completed laws are:
${(a^{m})}^{n} = a^{m n} and a^{- m} = \frac{1}{a^{m}}$
A power of a power multiplies the indices, and a negative index means one over the matching positive power.
What does a fractional index such as $a^{\frac{m}{n}}$ mean?
The denominator gives a root and the numerator gives a power: $a^{\frac{m}{n}} = \sqrt[n]{a^{m}} = {(\sqrt[n]{a})}^{m}$ .
Both orders give the same answer, so take the root first when that keeps the numbers smaller.
Why is $a^{0}$ equal to $1$ for any non-zero $a$ ?
Because dividing a power by itself gives $1$ , while the division law makes that same calculation $a^{m - m} = a^{0}$ .
The two answers must agree, so $a^{0} = 1$ ; the case $a = 0$ has to be excluded because $0^{0}$ has no value.
How do you simplify an expression such as $(3 x^{7}) \times (6 x^{4})$ ?
Deal with the numbers and the letters separately: multiply the coefficients, then apply the index law to the algebraic parts.
Here $3 \times 6 = 18$ and $x^{7} \times x^{4} = x^{11}$ , giving $18 x^{11}$ .
True or False?
If $a^{x} = a^{y}$ then $x = y$ , whatever the value of $a$ .
False.
This works only when the base $a$ is positive and not equal to $1$ .
If $a = 1$ then $1^{2} = 1^{5}$ while $2 \neq 5$ , so the indices need not match at all.
When solving $8^{x} = \frac{1}{4}$ by comparing indices, what is the first thing to do?
Rewrite both sides as powers of the same base.
Here $8 = 2^{3}$ and $\frac{1}{4} = 2^{- 2}$ , so the equation becomes $2^{3 x} = 2^{- 2}$ , giving $3 x = - 2$ and $x = - \frac{2}{3}$ .
Complete the product and quotient laws of logarithms:
$\log_{a} x y = \log_{a} x______\log_{a} y and \log_{a} \frac{x}{y} = \log_{a} x______\log_{a} y$
The completed laws are:
$\log_{a} x y = \log_{a} x + \log_{a} y and \log_{a} \frac{x}{y} = \log_{a} x - \log_{a} y$
A product inside a logarithm becomes a sum of logarithms, and a quotient becomes a difference.
How can the power $k$ be moved out of $\log_{a} x^{k}$ ?
The power becomes a multiplier in front of the logarithm: $\log_{a} x^{k} = k \log_{a} x$ .
This holds for any value of $k$ , negative and fractional ones included, which is why $\log_{a} \frac{1}{x} = \log_{a} x^{- 1} = - \log_{a} x$ .
True or False?
$\log_{a} (x + y) = \log_{a} x + \log_{a} y$
False.
Testing it with numbers settles the matter: $\log_{10} (1 + 99) = \log_{10} 100 = 2$ , whereas $\log_{10} 1 + \log_{10} 99 = 0 + 1.9956 \dots$
The same warning applies to $\log_{a} (x - y)$ , which is not $\log_{a} x - \log_{a} y$ .
What are the values of $\log_{a} a$ and $\log_{a} 1$ ?
$\log_{a} a = 1$ and $\log_{a} 1 = 0$ .
Both read straight off the definition of a logarithm, since $a^{1} = a$ and $a^{0} = 1$ .
Which index law does $\log_{a} x y = \log_{a} x + \log_{a} y$ correspond to, and why?
The product law of logarithms matches $a^{m} \times a^{n} = a^{m + n}$ .
A logarithm is an index, so multiplying two numbers means adding their indices, and that addition is exactly what the logarithm law records.
Why would you change the base of a logarithm?
Because the laws of logarithms only combine logarithms that share a base, so logarithms with different bases have to be rewritten to a common one first.
Choosing the new base well can also make a problem solvable without a calculator, as when $\log_{4} x$ and $\log_{32} x$ are both rewritten to base $2$ .
Complete the change of base formula:
$\log_{a} x = \frac{\log_{b}______}{\log_{b}______}$
The completed formula is:
$\log_{a} x = \frac{\log_{b} x}{\log_{b} a}$
The number goes on top and the original base goes underneath, while the new base $b$ can be anything convenient.
Show how $\log_{a} b = \frac{1}{\log_{b} a}$ follows from the change of base formula.
Change $\log_{a} b$ to base $b$ :
$\log_{a} b = \frac{\log_{b} b}{\log_{b} a} = \frac{1}{\log_{b} a}$
The numerator collapses to $1$ because a logarithm of its own base is always $1$ .
Define exponential equation.
An exponential equation is an equation in which the unknown appears in an index, such as $5^{x} = 27$ or $4^{x} - 3 (2^{x}) + 2 = 0$ .
What do you do to both sides of $5^{x} = 27$ when the answer cannot be spotted?
Take logarithms of both sides, then use the power law to bring $x$ down out of the index.
A logarithm to any base will do, though $\ln$ is the usual choice: $\ln (5^{x}) = \ln 27$ becomes $x \ln 5 = \ln 27$ .
Complete the exact solution of $5^{x} = 27$ , after taking logarithms of both sides:
$x \ln 5 = \ln 27 so x = \frac{\ln______}{\ln______}$
The completed solution is:
$x \ln 5 = \ln 27 so x = \frac{\ln 27}{\ln 5}$
Dividing by $\ln 5$ leaves the logarithm of the answer on top and the logarithm of the base underneath.
True or False?
$x = \frac{\ln 20}{\ln 3}$ is an exact solution of $3^{x} = 20$ .
True.
A quotient of two logarithms is an exact value, in just the same way that $\frac{1}{3}$ or $\sqrt{2}$ is exact.
A calculator turns it into $2.726833 \dots$ , which is a rounded version of the same number and so is less accurate.
How can you tell that $4^{x} - 3 (2^{x}) + 2 = 0$ is really a quadratic?
Because $4^{x} = {(2^{2})}^{x} = {(2^{x})}^{2}$ , so writing $y = 2^{x}$ turns the equation into $y^{2} - 3 y + 2 = 0$ .
Look for one power that is the square of another; $e^{2 x} = {(e^{x})}^{2}$ works in exactly the same way.
How do you rewrite $2^{x + 1}$ so that a hidden quadratic can be seen?
Split the index using the multiplication law: $2^{x + 1} = 2^{1} \times 2^{x} = 2 \times 2^{x}$ .
The term is then a multiple of $2^{x}$ rather than a new power, so it fits the quadratic alongside the other terms.
After solving a hidden quadratic in $2^{x}$ , what still has to be done?
Each value found for $2^{x}$ has to be turned back into a value of $x$ , using logarithms.
Discard any value that makes $2^{x}$ zero or negative, because no index can produce one: $2^{x} = - 5$ has no solution at all.

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