True or False? When changing the base of a logarithm, you may use any positive base you like.

True. The formula holds for any positive new base, so you pick whichever makes the arithmetic easiest. The answer comes out the same whichever you choose, since the original logarithm has one fixed value.

International A Level (IAL)MathsEdexcelPure 2FlashcardsExponentials & LogarithmsLaws of Logarithms

Laws of Logarithms (Edexcel International A Level (IAL) Maths: Pure 2): Flashcards

Exam code: YMA01

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FrontExponential Functions
Define exponential function.

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Cards in this collection (29)

Define exponential function.
A function of the form $y = a^{x}$ , where the variable sits in the power and $a > 0$ .
It is the variable's position that makes it exponential, not the size of the base.
What do all graphs of the form $y = a^{x}$ have in common?
They all pass through $(0, 1)$ , because $a^{0} = 1$ whatever $a$ is.
The $x$ -axis is an asymptote, so the curve approaches it without ever reaching it.
$y = a^{x}$ shows exponential growth when $a______$ , and exponential decay when $______< a <______$ .
It shows growth when $a > 1$ , and decay when $0 < a < 1$ .
A base bigger than $1$ multiplies up at every step, while a base between $0$ and $1$ multiplies down.
True or False?
$y = 1^{x}$ is an exponential function.
False.
$1^{x} = 1$ for every value of $x$ , so the graph is just the horizontal line $y = 1$ .
That is precisely why $a = 1$ is excluded: nothing grows and nothing decays.
For $x > 0$ , is $3^{x}$ above or below $2^{x}$ , and what happens for $x < 0$ ?
Above for $x > 0$ , and below for $x < 0$ .
The two curves meet at $(0, 1)$ , which every exponential graph passes through, so that is where the ordering swaps.
Why must the base of an exponential function be positive?
Because a negative base gives no real value for fractional powers.
${(- 4)}^{\frac{1}{2}}$ does not exist as a real number, so the graph would have gaps instead of being a smooth curve.
Define logarithm.
A logarithm answers the question what power do you raise the base to?: $\log_{b} a = x$ means exactly the same as $a = b^{x}$ .
The number $b$ is called the base.
How should you read the statement $\log_{p} q = r$ in words?
"The power you raise $p$ to, to get $q$ , is $r$ ."
Saying it that way every time turns the notation into a question that usually answers itself.
True or False?
${(\log x)}^{2}$ and $\log x^{2}$ mean the same thing.
False.
${(\log x)}^{2}$ squares the answer, while $\log x^{2}$ squares $x$ before the logarithm is taken.
For $x = 10$ the first gives $1$ and the second gives $2$ .
Complete these two results:
$\log_{a} a^{x} =______$
$a^{\log_{a} x} =______$
The completed results are:
$\log_{a} a^{x} = x$
$a^{\log_{a} x} = x$
Taking a logarithm and raising to a power are inverse operations, so doing both in either order leaves you where you started.
How do you evaluate $\log_{2} 32$ without a calculator?
Ask what power of $2$ gives $32$ .
Since $2^{5} = 32$ , the answer is $5$ .
Why can you not take the logarithm of a negative number?
Because no power of a positive base ever produces a negative result.
$\log_{b} a = x$ requires $b^{x} = a$ , and $b^{x}$ is positive for every value of $x$ when $b$ is positive.
What does $\log x$ mean when no base is written?
Base $10$ , so $\log x$ is short for $\log_{10} x$ .
It is sometimes written $\lg x$ instead, since $10$ is a common enough base that it is usually left off altogether.
Complete the three laws of logarithms:
$\log_{a} x y = \log_{a} x +______$
$\log_{a} (\frac{x}{y}) = \log_{a} x -______$
$\log_{a} x^{k} =______$
The completed laws are:
$\log_{a} x y = \log_{a} x + \log_{a} y$
$\log_{a} (\frac{x}{y}) = \log_{a} x - \log_{a} y$
$\log_{a} x^{k} = k \log_{a} x$
Multiplying becomes adding, dividing becomes subtracting, and a power becomes a multiplier.
True or False?
$\log (x + y) = \log x + \log y$
False.
There is no law that breaks up the logarithm of a sum: the laws apply to products, quotients and powers only.
With $x = y = 10$ the left-hand side is $\log 20$ , roughly $1.3$ , while the right-hand side is $2$ .
What are $\log_{a} a$ and $\log_{a} 1$ equal to?
$\log_{a} a = 1$ and $\log_{a} 1 = 0$ .
Both follow straight from what a logarithm means, since $a^{1} = a$ and $a^{0} = 1$ .
Why does $\log_{a} (\frac{1}{x})$ come out as $- \log_{a} x$ ?
Because $\frac{1}{x} = x^{- 1}$ , and the power law brings that $- 1$ out to the front.
So a reciprocal inside a logarithm always turns into a minus sign outside it.
You solve a logarithmic equation and get two values of $x$ . What must you do before giving both of them as answers?
Substitute each one back and reject any that makes the inside of a logarithm zero or negative.
In $\log (x + 2)$ , for instance, any value at or below $x = - 2$ has to go.
Complete the change of base formula, which is given in the formulae booklet:
$\log_{a} x = \frac{\log_{b} x}{______}$
The completed formula is:
$\log_{a} x = \frac{\log_{b} x}{\log_{b} a}$
The original base $a$ drops into the denominator, and the new base $b$ is written the same way in both places.
True or False?
When changing the base of a logarithm, you may use any positive base you like.
True.
The formula holds for any positive new base, so you pick whichever makes the arithmetic easiest.
The answer comes out the same whichever you choose, since the original logarithm has one fixed value.
When is the change of base formula actually worth using?
When two logarithms in the same problem have different bases, so one has to be rewritten to match the other.
It also helps where the two numbers are powers of a common number, as in $\log_{4} 8 = \frac{\log_{2} 8}{\log_{2} 4} = \frac{3}{2}$ .
What does the change of base formula give when the new base is $x$ itself?
$\log_{a} x = \frac{1}{\log_{x} a}$ , which swaps the two numbers round.
The numerator becomes $\log_{x} x$ , and that is $1$ .
Define exponential equation.
An exponential equation is one in which the unknown appears as a power.
$2^{x} = 8$ and $5^{2 x - 1} = 7$ are both exponential equations.
How do you solve $a^{f (x)} = b$ when $b$ can be written as a power of $a$ ?
Write $b$ as $a^{k}$ so that both sides share a base, then equate the powers and solve $f (x) = k$ .
No logarithms are needed at all once the bases match.
How do you solve $a^{f (x)} = b$ when $b$ is not a power of $a$ ?
Take logarithms of both sides, which gives $f (x) = \log_{a} b$ if you work in base $a$ .
Any base will do, because the power law brings the exponent down to the front whichever one you pick.
True or False?
$3 \times 2^{x}$ can be rewritten as $6^{x}$ .
False.
The multiplier sits outside the power, so it cannot be absorbed into the base: $6^{x}$ would mean $3^{x} \times 2^{x}$ .
At $x = 2$ the first expression is $12$ and the second is $36$ .
Complete the split, using the index law in reverse:
$2^{x + 3} = 2^{x} \times______$
The completed split is:
$2^{x + 3} = 2^{x} \times 2^{3} = 8 \times 2^{x}$
Splitting a power like this is what turns a three-term exponential equation into a quadratic in $2^{x}$ .
An equation contains $p \times a^{f (x)}$ , where $p$ is a constant. How do logarithms deal with the $p$ ?
Taking logarithms turns the product into a sum: $\log (p \times a^{f (x)}) = \log p + f (x) \log a$ .
$\log p$ is just a number, so what is left rearranges like any other equation.
How do you solve $a^{f (x)} = b^{g (x)}$ when neither base is a power of the other?
Take logarithms of both sides to the same base, then bring both exponents down to give $f (x) \log a = g (x) \log b$ .
Both logarithms are constants, so the equation left behind is a straightforward one to rearrange for $x$ .

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