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Laws of Logarithms (Cambridge (CIE) A Level Maths: Pure 3): Flashcards

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FrontLaws of Logarithms
Complete the three laws of logarithms:
$\log_{a} x y =______$
$\log_{a} (\frac{x}{y}) =______$
$\log_{a} x^{k} =______$

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Cards in this collection (16)

Complete the three laws of logarithms:
$\log_{a} x y =______$
$\log_{a} (\frac{x}{y}) =______$
$\log_{a} x^{k} =______$
The completed laws are:
$\log_{a} x y = \log_{a} x + \log_{a} y$
$\log_{a} (\frac{x}{y}) = \log_{a} x - \log_{a} y$
$\log_{a} x^{k} = k \log_{a} x$
In words: multiplying becomes adding, dividing becomes subtracting, and a power becomes a multiplier out in front.
Where do the laws of logarithms come from?
They are the laws of indices in disguise, because a logarithm is itself a power.
Behind the product law sits $a^{x} \times a^{y} = a^{x + y}$ , and behind the power law sits ${(a^{x})}^{y} = a^{x y}$ .
True or False?
$\log (x + y) = \log x + \log y$
False.
There is no law that breaks up the logarithm of a sum: only a product, a quotient or a power can be split.
Taking $x = y = 10$ in base $10$ gives $\log 20 \approx 1.301$ on the left, against $1 + 1 = 2$ on the right.
What are $\log_{a} a$ and $\log_{a} 1$ equal to?
The first is $1$ and the second is $0$ .
Both come straight from reading a logarithm as a power: you raise $a$ to the power $1$ to get $a$ , and to the power $0$ to get $1$ .
Do the laws of logarithms apply to $\ln$ ?
Yes, because $\ln x$ is simply $\log_{e} x$ , a logarithm whose base happens to be $e$ .
Nothing in any of the laws depends on which base is used, so every one of them applies to $\ln$ unchanged.
What does $\log_{a} (\frac{1}{x})$ simplify to?
It simplifies to $- \log_{a} x$ .
Writing $\frac{1}{x}$ as $x^{- 1}$ makes it $\log_{a} x^{- 1}$ , and the power law then brings the $- 1$ out to the front.
True or False?
A logarithm can come out negative.
True.
For example $\log_{10} 0.01 = - 2$ , because $10^{- 2} = 0.01$ .
It is the number you take the logarithm of that has to be positive; the logarithm itself can be any real number.
You have reduced a logarithmic equation to $\log_{2} (\frac{x^{2}}{x + 6}) = 3$ . What is the next move?
Undo the logarithm by making each side a power of the base, giving $\frac{x^{2}}{x + 6} = 2^{3} = 8$ .
That removes the logarithm altogether and leaves an ordinary equation, here a quadratic once you multiply up by $x + 6$ .
Why must you check the solutions of a logarithmic equation?
Because $\log_{a} x$ is only defined when $x > 0$ , so any value making a logarithm in the original equation take a negative or zero argument must be rejected.
Solving $2 \log_{2} x = 3 + \log_{2} (x + 6)$ gives $x = 12$ and $x = - 4$ , and $- 4$ goes because $\log_{2} (- 4)$ does not exist.
Define an exponential equation.
An equation in which the unknown is a power.
For example $3^{3 x} - 4 = 9^{x} + 5$ .
When can an exponential equation be solved without using logarithms at all?
When both sides can be written as powers of the same base, because then the powers themselves must be equal.
$5^{2 x} = 125 \Rightarrow 5^{2 x} = 5^{3}$ , so $2 x = 3$ and $x = \frac{3}{2}$ .
How do you solve an exponential equation whose two sides cannot be written as powers of the same base?
Take logarithms of both sides, then use $\log_{a} x^{k} = k \log_{a} x$ to bring each power down as a multiplier, then rearrange for $x$ .
Natural logarithms are usual, and exact answers are normally left in terms of $\ln$ .
True or False?
$\frac{\ln 42}{\ln 6} = \ln 7$
False.
Logarithms cannot be cancelled or divided like that. $\frac{\ln 42}{\ln 6}$ is one number divided by another and does not simplify.
It is subtraction that combines them: $\ln 42 - \ln 6 = \ln 7$ .
How do you spot and handle a hidden quadratic in an exponential equation?
Look for one exponential term that is the square of another: $4^{x} = {(2^{x})}^{2}$ and $e^{2 x} = {(e^{x})}^{2}$ .
Then substitute, for example, $y = e^{x}$ , solve the resulting quadratic, then solve $e^{x} = y$ for each root.
Complete the rearrangement into quadratic form:
$21 e^{x} - 4 = 5 e^{2 x}$
becomes
$5 {(______)}^{2} -______(e^{x}) +______= 0$
The rearrangement gives:
$5 {(e^{x})}^{2} - 21 (e^{x}) + 4 = 0$
which factorises as $(5 e^{x} - 1) (e^{x} - 4) = 0$ .
An answer must be given as $x = \frac{\ln p}{\ln q}$ with $p$ and $q$ integers. How do you get there from $x = \frac{6 \ln 2 + \ln 3}{2 \ln 3 + \ln 2}$ ?
Use the laws of logarithms in reverse to collapse the top and bottom into single logarithms.
$6 \ln 2 = \ln 64$ , so the top is $\ln 64 + \ln 3 = \ln 192$ . The bottom is $\ln 9 + \ln 2 = \ln 18$ , giving $x = \frac{\ln 192}{\ln 18}$ .

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