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Laws of Logarithms (Edexcel A Level Maths: Pure): Flashcards

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FrontLaws of Logarithms
Complete the three laws of logarithms:
$\log_{a} x + \log_{a} y =______$
$\log_{a} x - \log_{a} y =______$
$k \log_{a} x =______$

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Complete the three laws of logarithms:
$\log_{a} x + \log_{a} y =______$
$\log_{a} x - \log_{a} y =______$
$k \log_{a} x =______$
The completed laws are:
$\log_{a} x + \log_{a} y = \log_{a} (x y)$
$\log_{a} x - \log_{a} y = \log_{a} (\frac{x}{y})$
$k \log_{a} x = \log_{a} x^{k}$
These are written as the exam board writes them, combining two logarithms into one. You will just as often use them in reverse, to split one logarithm into two.
True or False?
For any positive $x$ and $y$ , $\log (x + y) = \log x + \log y$ .
False.
There is no law for the logarithm of a sum.
The law is about a product: $\log x + \log y = \log (x y)$ . Treating a sum this way is one of the most heavily penalised errors in the topic.
Why is $\log_{a} a = 1$ and $\log_{a} 1 = 0$ ?
A logarithm answers the question "what power do you raise $a$ to?".
To get $a$ you need power 1. To get 1 you need power 0, since $a^{0} = 1$ .
Why does $\log_{a} \frac{1}{x} = - \log_{a} x$ ?
Because $\frac{1}{x} = x^{- 1}$ , and the power law with $k = - 1$ gives $\log_{a} x^{- 1} = - \log_{a} x$ .
The specification expects negative and fractional values of $k$ , including $k = - \frac{1}{2}$ .
How do you write $3 \log_{2} (2 x + 3) + \log_{2} 5 - 2 \log_{2} (x + 1)$ as a single logarithm?
Clear the multipliers with the power law first, then combine with the product and quotient laws:
$\log_{2} (2 x + 3)^{3} + \log_{2} 5 - \log_{2} (x + 1)^{2} = \log_{2} \frac{5 (2 x + 3)^{3}}{(x + 1)^{2}}$
Define ln.
The natural logarithm: a logarithm whose base is the constant $e$ , so $\ln x \equiv \log_{e} x$ .
It is a function, not a number.
What are $e^{\ln x}$ and $\ln (e^{x})$ equal to, and why?
Both are equal to $x$ .
$e^{x}$ and $\ln x$ are inverse functions, so each undoes the other. This is what lets you remove an $e$ or an $\ln$ from an equation.
What must you check after solving an equation that contains logarithms?
That every solution is valid. You can only take the logarithm of a positive number, so $\log (x + k)$ only exists when $x > - k$ .
Marks are lost for failing to reject a solution that makes a logarithm undefined.
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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