Ask: "3 to what power = 27?"
3¹=3, 3²=9, 3³=27 ✓
So \(\log_3 27 = 3\). Answer: C
2
BASIC
⭐
Which expression equals \(\log_5 1\)?
💡 Explanation
Rule: \(\log_b 1 = 0\) for any valid base \(b\).
Reason: \(b^0 = 1\) always. So \(\log_5 1 = 0\). Answer: B
3
TRICKY ⚠️
⭐⭐
Convert to exponential form: \(\log_4 x = 3\)
💡 Explanation
Use B-E-A rule: base=4, exponent=3, answer=x
\(\log_4 x = 3 \Rightarrow 4^3 = x \Rightarrow x = 64\)
Watch out! Many mix up base and exponent. Answer: C
4
TRICKY ⚠️
⭐⭐
What is \(\log_{10} 0.01\)?
💡 Explanation
\(0.01 = \frac{1}{100} = 10^{-2}\)
So \(\log_{10}(10^{-2}) = -2\).
Key insight: logs of numbers between 0 and 1 are NEGATIVE! Answer: D
2
📗 Unit 2: Logarithm Rules (Laws)
PRODUCT RULE
\[\log_b(mn) = \log_b m + \log_b n\]
QUOTIENT RULE
\[\log_b\!\left(\tfrac{m}{n}\right) = \log_b m - \log_b n\]
POWER RULE
\[\log_b(m^k) = k\cdot\log_b m\]
CHANGE OF BASE
\[\log_b a = \frac{\log a}{\log b} = \frac{\ln a}{\ln b}\]
Which is the correct simplification of \(\log_5(x^3)\)?
💡 Explanation
Power rule: the exponent moves to the FRONT as a coefficient.
\(\log_5(x^3) = 3 \cdot \log_5 x\)
Most common mistake: writing \((\log_5 x)^3\) — the power applies to \(x\), NOT to the log! Answer: C
8
TRICKY ⚠️
⭐⭐⭐
Evaluate: \(\log_8 2\) using change of base
💡 Explanation
\(\log_8 2 = \dfrac{\log 2}{\log 8} = \dfrac{\log 2}{\log 2^3} = \dfrac{\log 2}{3 \log 2} = \dfrac{1}{3}\)
OR: ask "8 to what power = 2?" → \(8^{1/3} = \sqrt[3]{8} = 2\). Answer: B
3
📙 Unit 3: Logarithmic Equations
🔑 SAME BASE = SAME ARGUMENT
If \(\log_b A = \log_b B\) then \(A = B\).
If \(\log_b A = k\) then \(A = b^k\). ⚠️ Always CHECK: argument must be > 0 !
After solving, ALWAYS check that all arguments of log are POSITIVE. If not, discard that solution!
9
EQUATION
⭐⭐
Solve: \(\log_3 x = 4\)
💡 Explanation
\(\log_3 x = 4 \Rightarrow x = 3^4 = 81\). Answer: B
10
EQUATION
⭐⭐⭐
Solve: \(\log_2 x + \log_2(x-2) = 3\)
💡 Explanation
Product rule: \(\log_2[x(x-2)] = 3 \Rightarrow x(x-2) = 8\)
\(x^2 - 2x - 8 = 0 \Rightarrow (x-4)(x+2) = 0\)
So \(x=4\) or \(x=-2\). But \(x=-2\) makes \(\log_2(-2)\) undefined! ✗
Answer: \(x = 4\). Answer: C
11
EQUATION
⭐⭐
Solve: \(\log_5(2x-1) = \log_5(x+4)\)
💡 Explanation
Same base → same argument: \(2x-1 = x+4\)
\(x = 5\). Check: \(2(5)-1=9>0\) ✓ Answer: D
12
TRICKY ⚠️
⭐⭐⭐
Which value of \(x\) makes \(\log_3(x^2 - x)\) undefined?
💡 Explanation
Need \(x^2 - x > 0\) for log to be defined.
At \(x=0\): \(0^2 - 0 = 0\). Log of 0 is UNDEFINED! ✗
At \(x=1\): also 0. But from given options, x=0 is the clear answer. Answer: C
4
📒 Unit 4: Logarithmic Functions & Graphs
🔑 LOG ↔ EXP are MIRRORS (y = x)
The graph of \(y = \log_b x\) is the REFLECTION of \(y = b^x\) across the line \(y=x\). Domain of log: x > 0 | Range: all real numbers (−∞, +∞)
📈 INTERACTIVE GRAPH REFERENCE
y = log₂x (b > 1, increasing)
y = log₀.₅x (0 < b < 1, decreasing)
b > 1 (e.g. log₂x)
✅ Increasing (goes up)
✅ Passes through (1, 0)
✅ Passes through (b, 1)
✅ x-axis is vertical asymptote approach
0 < b < 1 (e.g. log₀.₅x)
❌ Decreasing (goes down)
✅ Passes through (1, 0)
✅ Domain: x > 0 only
⚠️ Still has same shape, just flipped!
13
GRAPH
⭐⭐
The graph of \(y = \log_2 x\) passes through which point?
💡 Explanation
\(\log_b 1 = 0\) for ANY base b. So \(y = \log_2(1) = 0\) → point (1, 0).
Note: (0, ?) is impossible since log(0) is undefined! Answer: C
14
GRAPH
⭐⭐
What is the domain of \(f(x) = \log_3(x - 2)\)?
💡 Explanation
The argument (inside the log) must be > 0:
\(x - 2 > 0 \Rightarrow x > 2\)
The graph is shifted RIGHT 2 units. Answer: C
15
GRAPH
⭐⭐⭐
The graph of \(y = \log_{0.5} x\) is _____ compared to \(y = \log_2 x\).
💡 Explanation
\(\log_{0.5} x = \log_{2^{-1}} x = -\log_2 x\)
So it's \(-1 \times \log_2 x\) → reflected across the x-axis (flipped upside down). Answer: B
16
GRAPH
⭐⭐⭐
Where is the vertical asymptote of \(f(x) = \log_2(x + 3)\)?
💡 Explanation
Asymptote is where the argument = 0:
\(x + 3 = 0 \Rightarrow x = -3\)
The graph shifts LEFT 3, so asymptote moves to \(x = -3\). Answer: C
5
🏆 Unit 5: Mixed Challenge Problems
🔑 "ANTI-LOG" = EXPONENTIATE
To undo a log, raise the base to both sides.
\(\log_b(\text{something}) = k \;\Rightarrow\; \text{something} = b^k\) Think of log and exponent as "opposites" that cancel each other out.
17
TRICKY ⚠️
⭐⭐⭐
Evaluate: \(2^{\log_2 5}\)
💡 Explanation
Key identity: \(b^{\log_b x} = x\)
So \(2^{\log_2 5} = 5\).
The base and the log base cancel! Answer: B
18
EQUATION
⭐⭐⭐
Solve for \(x\): \(\log x + \log(x+3) = 1\) (base 10)
As \(x \to 0^+\), what happens to \(y = \log_2 x\)?
💡 Explanation
As x gets very small (approaching 0 from the right), the log goes to negative infinity.
Try: \(\log_2(0.5) = -1\), \(\log_2(0.25) = -2\), \(\log_2(0.001) \approx -10\)
The y-axis (\(x=0\)) is a vertical asymptote going to \(-\infty\). Answer: D
20
TRICKY ⚠️
⭐⭐⭐⭐
If \(\log_a 2 = p\) and \(\log_a 3 = q\), express \(\log_a 12\) in terms of \(p\) and \(q\).
💡 Explanation
\(12 = 4 \times 3 = 2^2 \times 3\)
\(\log_a 12 = \log_a(2^2 \cdot 3) = \log_a 2^2 + \log_a 3 = 2\log_a 2 + \log_a 3 = 2p + q\)
Decompose the number into prime factors first! Answer: C