A LevelMathsCambridge (CIE)Pure 3FlashcardsLogarithms & ExponentialsLaws of Logarithms

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

Exam code: 9709

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FrontExponential Equations
Define an exponential equation.

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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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