IB Mathematics AA SL Exponents and logarithms Study Notes
IB Mathematics AA SL Exponents and logarithms Study Notes Offer a clear explanation of Exponents and logarithms , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Exponents and logarithms
Exponents and Logarithms
Exponents
Exponents are a way to express repeated multiplication of a number. The exponent indicates how many times the base number is multiplied by itself.
For example:
- 23 = 2 × 2 × 2 = 8
Some important properties of exponents include:
- am × an = am+n
- am / an = am-n
- (am)n = amn
- a0 = 1 (where a ≠ 0)
Logarithms
Logarithms are the opposite of exponents. They help to determine the exponent to which a base must be raised to produce a given number.
There are some laws to remember:
- logα α = 1
- logα 1 = 0
- logα ak = k logα a
- logα (xy) = logα x + logα y (same base only)
- logα (x/y) = logα x – logα y (same base only)
It’s a way of rewriting an exponent.
Examples:
- logx 64 = 6
- log2 x = 4
Solutions:
- x6 = 64 → x = 2
- 2x = 16 → x = 4
Change in Base Formula
The change in base formula is given by:
\( \log_b a = \frac{\log_c b}{\log_c a} \)
Combine / Change of Base
Using the change of base property, we can combine logarithms:
\( \log_4 5 + \log_4 16 = \log_4 (5 \times 16) = \log_4 80 \)
Rewriting Logarithms in Terms of x and y
Some questions will require you to rewrite logarithms in terms of \(x\) and \(y\). In this case, use logarithms accordingly.
Natural Logs
The value of \(e\) is approximately:
\( e \approx 2.718281828459045… \)
It sits between 2 and 3, and it is the base of the natural logarithm. The gradient of \(e^x\) is \(e\), which is the only function to do so.
How to Use a Calculator
- For \( \log_2 x \), use the “log” function.
- For \( \log_e x \), use the “ln” function.
IB Mathematics AA SL Exponents and logarithm Exam Style Worked Out Questions
Question
Solve the equation \(2 – {\log _3}(x + 7) = {\log _{\tfrac{1}{3}}}2x\) .
▶️Answer/Explanation
Markscheme
\({\log _3}\left( {\frac{9}{{x + 7}}} \right) = {\log _3}\frac{1}{{2x}}\) M1M1A1
Note: Award M1 for changing to single base, M1 for incorporating the 2 into a log and A1 for a correct equation with maximum one log expression each side.
\(x + 7 = 18x\) M1
\(x = \frac{7}{{17}}\) A1
[5 marks]
Question
Solve the equation \({4^{x – 1}} = {2^x} + 8\).
▶️Answer/Explanation
Markscheme
\({2^{2x – 2}} = {2^x} + 8\) (M1)
\(\frac{1}{4}{2^{2x}} = {2^x} + 8\) (A1)
\({2^{2x}} – 4 \times {2^x} – 32 = 0\) A1
\(({2^x} – 8)({2^x} + 4) = 0\) (M1)
\({2^x} = 8 \Rightarrow x = 3\) A1
Notes: Do not award final A1 if more than 1 solution is given.
[5 marks]