[qdeck ” ]
[h] IB Mathematics AA HL Flashcards- Exponents and logarithms
[q] Exponents and logarithms
[a]
As you should know, an exponent is a way of expressing repeated multiplication, i.e. \( x^c = x \cdot x \cdot x \cdot x \cdot x \). In this section, we will explore: the products & division of powers of \( x \), powers of powers, negative exponents, and the ‘inverse’ of exponents → logarithms.
[q]
MULTIPLICATION // We can discover these rules through simple examples:
If we simplify \( x^3 \cdot x^4 \), we get \( (x \cdot x \cdot x) \cdot (x \cdot x \cdot x \cdot x) \) which is simply \( x^7 \) (or \( x^{3 + 4} \)). Hence, we get the rule:
\[
x^a \cdot x^b = x^{a + b} \quad \text{Not in F.B.}
\]
[a]
E.G.1 // Simplify \( y^2 \cdot y^{11} \):
\[
y^{2 + 11} = y^{13}
\]
[q]
E.G.2 // Simplify \( x^6 \cdot x^{-4} \):
\[
x^{6 + (-4)} = x^2
\]
[a]
DIVISION // Through simplification of fractions, we can discover a rule:
If we take \( x^5 ÷ x^3 \):
\[
x^5 ÷ x^3 = \frac{x \cdot x \cdot x \cdot x \cdot x}{x \cdot x \cdot x} = x \cdot x = x^2 \quad \text{(or } x^{5 – 3})
\]
[q]
Hence, we get:
\[
x^a ÷ x^b = x^{a – b} \quad \text{Not in F.B.}
\]
E.G.3 // Calc. \( 3^{13} ÷ 3^{11} \):
\[
3^{13 – 11} = 3^2 = 9
\]
[a]
E.G.4 // Simplify \( x^6 ÷ x^5 y = x^{6 – 5} = x y^4 \)
NEGATIVES // If we take the expression \( x^3 \), we know this can also be written as \( 1 \cdot x \cdot x \cdot x \). So \( x^{-3} = \frac{1}{x^3} \). Hence, we get:
\[
x^{-n} = \frac{1}{x^n} \quad \text{Not in F.B.}
\]
E.G.5 // Express \( \frac{1}{16} \) as a power of 2:
\[
\frac{1}{16} = \frac{1}{2^4} = 2^{-4}
\]
[q]
E.G.6 // \[ Simplify \( 27^{\frac{1}{3}} \cdot 27^{-1} = \frac{27^{\frac{1}{3}}}{27} = \frac{3}{27} = \frac{1}{9}\]
OTHER RULES // Also note:
\[
(ab)^n = a^n \cdot b^n \quad \text{&} \quad \left( \frac{a}{b} \right)^n = \frac{a^n}{b^n} \quad \text{Neither in F.B.}
\]
[a]
LOGARITHMS // We have always used inverse operations of the basic operations to help solve equations. Until now, we have not learned what operation to use when solving \( 2^x = 4096 \) or \( 3^x = \frac{1}{81} \).
[q]
But we can actually use the following:
\[
a^x = b \leftrightarrow x = \log_a b \quad \text{In F.B.} \quad \text{where } a > 0 \quad \text{& } \ln x = \log_e x \quad b > 0 \quad a ≠ 1 \quad \log_{10} x = \log x
\]
This gives us a helpful notation, or just a simple calculation to type into a GDC.
[a]
E.G.7 // Find \( \log_6 36 \):
Solve \( 6^x = 36 \), \( x = 2 \)
E.G.8 // Solve \( 2^x = 49.1 \):
\( \log_2 49.1 = 5.62 \) (with GDC)
[q] Simple Proofs
1// This section will deal with simple deductive proofs, both numerical and algebraic. The majority of these will be ‘left-hand side’ to ‘right-hand side’ proofs → showing the LHS does indeed equal the RHS.
[a]
NOTATION// One thing you may come across with these proofs is an identity or equivalence symbol written with a ≡. This is different to an =, because an identity means ‘is always equal to’, so it only really should be used for an algebraic identity that holds for all x. An = symbol refers just to that unique situation. i.e.:
\[
\frac{1}{12} + \frac{1}{4} = \frac{1}{3}
\]
but
\[
\frac{1}{x^2 + x} + \frac{1}{x + 1} \equiv \frac{1}{x}
\]
[q]
1// This fact is broadly ignored across Maths, even in IB.
METHOD// If you need to show that the LHS ≡ RHS, then do not start manipulating both sides simultaneously. Just change the LHS until it equals the RHS.
E.G.1 // Show that
\[
\frac{1}{x^2 + x} + \frac{1}{x + 1} \equiv \frac{1}{x}
\]
\[
\text{LHS} = \frac{1}{x^2 + x} + \frac{1}{x + 1} = \frac{1}{x(x + 1)} + \frac{x}{x(x + 1)} = \frac{1 + x}{x(x + 1)} = \frac{1}{x} = \text{RHS}
\]
[a]
E.G.2 // Show \((x-3)^2 + 4 \equiv x^2 – 6x + 13\)
\[
\text{LHS} = (x-3)^2 + 4 = (x^2 – 6x + 9) + 4 = x^2 – 6x + 13 = \text{RHS}
\]
1// You could also do the same from RHS to LHS, as long as you just pick one.
1// You may need more trigonometry knowledge for E.G.3 [see topic 3]:
[q]
E.G.3 // Prove the identity: \(\frac{\cot(x)}{\csc(x)} \equiv \cos(x)\)
\[
\text{LHS} = \frac{\cot(x)}{\csc(x)} = \frac{\cos(x)/\sin(x)}{1/\sin(x)} = \cos(x) = \text{RHS}
\]
[a] More Logarithms
RATIONAL EXPONENTS// We will now look at how to evaluate powers in the form: \(a^{\frac{m}{n}}\), or even \(a^{m \cdot n}\).
Starting with \(a^{\frac{m}{n}}\) and specifically the example \(9^{\frac{1}{2}}\).
[q]
If we square this, we square to get 9, i.e., the square root of 9. So \(9^{\frac{1}{2}} = \sqrt{9}\). This can be extended to:
\[
a^{\frac{m}{n}} = \sqrt[n]{a^m}
\]
1// Then, if we take \(8^{\frac{2}{3}}\), we can do: \(8^{\frac{2}{3}} = (8^{\frac{1}{3}})^2 = (\sqrt[3]{8})^2 = 2^2 = 16\). So, we have:
\[
a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m
\]
[a]
E.G.1 // \(32^{\frac{5}{5}} = \sqrt[5]{32} = 2\)
E.G.2 // \(16^{\frac{5}{4}} = (16^{\frac{1}{4}})^5 = (\sqrt[4]{16})^5 = 2^5 = 1024\)
MORE LOGS// If we take the fact that \(\log(a)\) is equal to ‘the exponent of 10 that gives you \(a\)’, then it follows that if we take \(10^{\log(a)}\), then it must equal \(a\). In fact:
\[
x^{\log_x a} = a
\]
[q]
ADDITION// Using the formula above, we can derive another important formula:
If we first take \(10^{\log a} = a\), \(10^{\log b} = b\) and \(10^{\log ab} = ab\) (from above).
[a]
Starting with the third equation: \(10^{\log ab} = ab\) \(\text{sub. out } a \& b\) using exp. rules \(\rightarrow 10^{\log ab} = 10^{\log a + \log b}\) \(\text{equating exponents} \rightarrow \log ab = \log a + \log b\). Hence, the rule:
\[
\log_c(xy) = \log_c(x) + \log_c(y)
\]
[q]
1// Using the other exponent rules, and a similar derivation, we get:
\[
\log_c\left(\frac{x}{y}\right) = \log_c(x) – \log_c(y)
\]
&
\[
\log_c(x^m) = m\log_c(x)
\]
[a]
1// We can now have a look at some examples of using these log rules:
E.G.3 // Evaluate \(\log_4 (4 \times 9) = \log_6 (4 \times 9) = \log_6 36 = 2\)
E.G.4 // Evaluate \(\log_2 5^2 – \log_{10} \left(\frac{2}{10}\right) = \log_5 \left(\frac{2}{10}\right) = \log_5 \left(\frac{1}{5}\right) = -1\)
[q]
E.G.5 // Solve \(\log_3 27 + 3 \log_8 \left(\frac{1}{2}\right) – \log_{16} 4 = \log_4 x\):
\[
\begin{aligned}
3 + \log_8 \left(\frac{1}{2}\right)^3 – \frac{1}{2} &= \log_4 x \\
\log_8 \left(\frac{1}{8}\right) + \frac{5}{2} &= \log_4 x \\
-1 + \frac{5}{2} &= \log_4 x \\
\frac{3}{2} &= \log_4 x \\
4^{\frac{3}{2}} &= x \\
x &= \left(4^{\frac{1}{2}}\right)^3 = \left(\sqrt{4}\right)^3 = 2^3 = 8
\end{aligned}
\]
[a]
CHANGE OF BASE// A very helpful formula can be derived as follows:
Take \(y = \log_b x\) \(\rightarrow\) \(b^y = x\) \(\rightarrow\) take \(\log_a\) of both sides \(\rightarrow\) \(\log_a b^y = \log_a x\).
[q]
Use log power rule \(y \cdot \log_a b = \log_a x\) \(\rightarrow\) \(\log_b x \cdot \log_a b = \log_a x\), which rearranges to:
\[
\log_b x = \frac{\log_a x}{\log_a b}
\]
[a]
1// This is helpful for when the numerator or denominator is an integer, or if you already have a ‘log base a’ elsewhere in the expression/equation, so they can be combined using the other log rules:
E.G.6 // Solve \(32^x = 16\): \(x = \log_{32} 16 = \frac{\log_2 16}{\log_2 32} = \frac{4}{5}\)
[q]
E.G.7 // Evaluate \(\log_7 \left(\frac{1}{8}\right) \div \log_8 7\): \(\log_8 \left(\frac{1}{8}\right) = \log_8 \left(\frac{1}{8}\right) = -1\)
E.G.8 // Simplify \((\log_9 5)(\log_7 7)(\log_2 27)/(\log_2)\): Use change of base for each:
\[
\begin{aligned}
-\frac{\ln 5}{\ln 7} \times \frac{\ln 7}{\ln 3} \times \frac{\ln 27}{\ln 3} \times \log_3 27 = 3
\end{aligned}
\]
[a] Standard form:
A number in the format , \( \pm a \times10^k \) , where \(1\leqslant a\leqslant 10 \) and \(k\in Z\) (k is an integer)
[q] Standard Form – Solved Examples 1
The diameter of a spherical planet is 6 × 104 km .
(a) Write down the radius of the planet. [1]
The volume of the planet can be expressed in the form π(a × 10k) km3 where 1 ≤ a <10 and k ∈ \(\mathbb{Z}\)
(b) Find the value of a and the value of k .
[a] Answer – Solved Examples 1
(a) \(3 \times 10^4\)
(b) \(\frac{4}{3}\pi(3 \times 10^4)^3\)
\(=\frac{4}{3}\pi\times 27 \times 10^{12}\)
\(=\pi(3.6 \times 10^{13})(km)^3\)
Hence
\(a= 3.6 \) , \(k =13\)
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