IB Mathematics AA – Topic 1: Number & Algebra

The Seven Essential Exponent Laws:

• 1. Product Rule (Multiplication with Same Base): \(a^m \times a^n = a^{m+n}\)

  When multiplying powers with the same base, add the exponents

  Example: \(2^3 \times 2^5 = 2^{3+5} = 2^8 = 256\)

• 2. Quotient Rule (Division with Same Base): \(\displaystyle\frac{a^m}{a^n} = a^{m-n}\) (where \(a \neq 0\))

  When dividing powers with the same base, subtract the exponents

  Example: \(\displaystyle\frac{5^7}{5^3} = 5^{7-3} = 5^4 = 625\)

• 3. Power of a Power Rule: \((a^m)^n = a^{mn}\)

  When raising a power to another power, multiply the exponents

  Example: \((3^2)^4 = 3^{2 \times 4} = 3^8 = 6561\)

• 4. Power of a Product Rule: \((ab)^n = a^n b^n\)

  Distribute the exponent to each factor

  Example: \((2 \times 3)^3 = 2^3 \times 3^3 = 8 \times 27 = 216\)

• 5. Power of a Quotient Rule: \(\displaystyle\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}\) (where \(b \neq 0\))

  Distribute the exponent to numerator and denominator

  Example: \(\displaystyle\left(\frac{2}{3}\right)^4 = \frac{2^4}{3^4} = \frac{16}{81}\)

• 6. Zero Exponent Rule: \(a^0 = 1\) (where \(a \neq 0\))

  Any non-zero number raised to the power of zero equals 1

  Example: \(5^0 = 1\), \((-7)^0 = 1\), \((3.14)^0 = 1\)

• 7. Negative Exponent Rule: \(a^{-n} = \displaystyle\frac{1}{a^n}\) (where \(a \neq 0\))

  A negative exponent means "reciprocal"

  Example: \(2^{-3} = \displaystyle\frac{1}{2^3} = \frac{1}{8}\), also \(\displaystyle\frac{1}{5^{-2}} = 5^2 = 25\)

Fractional Exponents (Rational Exponents):

• Fractional Exponent (Root): \(a^{\frac{1}{n}} = \sqrt[n]{a}\)

  A fractional exponent with numerator 1 represents a root

  Example: \(27^{\frac{1}{3}} = \sqrt[3]{27} = 3\), \(16^{\frac{1}{4}} = \sqrt[4]{16} = 2\)

• General Fractional Exponent: \(a^{\frac{m}{n}} = \sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m\)

  The numerator is the power, the denominator is the root

  Example: \(8^{\frac{2}{3}} = \sqrt[3]{8^2} = \sqrt[3]{64} = 4\) or \(8^{\frac{2}{3}} = (\sqrt[3]{8})^2 = 2^2 = 4\)

⚠ Common Pitfalls:

• \((a + b)^n \neq a^n + b^n\) — This is a very common mistake! The power does NOT distribute over addition.
• \((a^m)^n = a^{mn}\) but \(a^{m^n} = a^{(m^n)}\) — These are different! Order matters with multiple exponents.
• When adding or subtracting powers: \(a^n + a^m\) cannot be simplified unless you factor (if possible).
• \(a^{-n} \neq -a^n\) — Negative exponent means reciprocal, not negative value.
• For fractional exponents: \(a^{\frac{m}{n}} \neq \frac{a^m}{a^n}\)

💡 Pro Tips:

• Always look for opportunities to express numbers with the same base (e.g., \(8 = 2^3\), \(9 = 3^2\)).
• When simplifying, work step by step and apply one law at a time.
• Convert fractional exponents to radical form (or vice versa) depending on which is easier to evaluate.
• Remember: \(1^n = 1\) for any \(n\), and \(0^n = 0\) for any positive \(n\).

Example 1: Simplifying Expressions with Exponents

Problem: Simplify the following expressions:

(a) \(3^4 \times 3^7 \div 3^5\)

(b) \(\displaystyle\frac{(2^3)^4 \times 2^{-5}}{2^6}\)

(c) \((27)^{\frac{2}{3}} \times (27)^{\frac{1}{3}}\)

Solution:

(a) \(3^4 \times 3^7 \div 3^5\)

Apply product rule and quotient rule:

\(= 3^{4+7} \div 3^5\)

\(= 3^{11} \div 3^5\)

\(= 3^{11-5}\)

\(= 3^6\)

\(= 729\)

(b) \(\displaystyle\frac{(2^3)^4 \times 2^{-5}}{2^6}\)

First apply power of a power rule:

\(= \displaystyle\frac{2^{3 \times 4} \times 2^{-5}}{2^6}\)

\(= \displaystyle\frac{2^{12} \times 2^{-5}}{2^6}\)

Apply product rule in numerator:

\(= \displaystyle\frac{2^{12+(-5)}}{2^6}\)

\(= \displaystyle\frac{2^7}{2^6}\)

Apply quotient rule:

\(= 2^{7-6}\)

\(= 2^1 = 2\)

(c) \((27)^{\frac{2}{3}} \times (27)^{\frac{1}{3}}\)

Apply product rule:

\(= 27^{\frac{2}{3} + \frac{1}{3}}\)

\(= 27^{\frac{3}{3}}\)

\(= 27^1\)

\(= 27\)

Alternative: \(27^{\frac{2}{3}} \times 27^{\frac{1}{3}} = (\sqrt[3]{27})^2 \times \sqrt[3]{27} = 3^2 \times 3 = 9 \times 3 = 27\)

Example 2: Working with Negative and Fractional Exponents

Problem: Simplify and evaluate:

(a) \(16^{-\frac{3}{4}}\)

(b) \(\displaystyle\left(\frac{8}{27}\right)^{-\frac{2}{3}}\)

(c) \(5^{-2} \times 5^3 \times 5^{-1}\)

Solution:

(a) \(16^{-\frac{3}{4}}\)

Apply negative exponent rule first:

\(= \displaystyle\frac{1}{16^{\frac{3}{4}}}\)

Convert fractional exponent to radical form:

\(= \displaystyle\frac{1}{\sqrt[4]{16^3}}\) or \(= \displaystyle\frac{1}{(\sqrt[4]{16})^3}\)

Since \(\sqrt[4]{16} = 2\) (because \(2^4 = 16\)):

\(= \displaystyle\frac{1}{2^3}\)

\(= \displaystyle\frac{1}{8}\)

(b) \(\displaystyle\left(\frac{8}{27}\right)^{-\frac{2}{3}}\)

Apply negative exponent rule (reciprocal):

\(= \displaystyle\left(\frac{27}{8}\right)^{\frac{2}{3}}\)

Apply power of a quotient rule:

\(= \displaystyle\frac{27^{\frac{2}{3}}}{8^{\frac{2}{3}}}\)

Evaluate each part: \(27^{\frac{2}{3}} = (\sqrt[3]{27})^2 = 3^2 = 9\)

And: \(8^{\frac{2}{3}} = (\sqrt[3]{8})^2 = 2^2 = 4\)

\(= \displaystyle\frac{9}{4}\)

\(= 2.25\) or \(\displaystyle\frac{9}{4}\)

(c) \(5^{-2} \times 5^3 \times 5^{-1}\)

Apply product rule (add all exponents):

\(= 5^{-2+3+(-1)}\)

\(= 5^{-2+3-1}\)

\(= 5^0\)

Apply zero exponent rule:

\(= 1\)

Example 3: Mixed Exponent Problems

Problem: Simplify the expression \(\displaystyle\frac{(x^2y^{-3})^3 \times (x^{-1}y^2)^2}{(xy)^{-4}}\), where \(x, y \neq 0\).

Solution:

Step 1: Apply power of a power rule to each bracket in numerator:

\((x^2y^{-3})^3 = x^{2 \times 3}y^{-3 \times 3} = x^6y^{-9}\)

\((x^{-1}y^2)^2 = x^{-1 \times 2}y^{2 \times 2} = x^{-2}y^4\)

Step 2: Multiply these results (product rule):

\(x^6y^{-9} \times x^{-2}y^4 = x^{6+(-2)}y^{-9+4} = x^4y^{-5}\)

Step 3: Simplify denominator using power of a product:

\((xy)^{-4} = x^{-4}y^{-4}\)

Step 4: Divide (quotient rule):

\(\displaystyle\frac{x^4y^{-5}}{x^{-4}y^{-4}} = x^{4-(-4)}y^{-5-(-4)}\)

\(= x^{4+4}y^{-5+4}\)

\(= x^8y^{-1}\)

Step 5: Convert negative exponent to positive form:

\(= \displaystyle\frac{x^8}{y}\)

The Fundamental Relationship:

The Six Essential Logarithm Laws:

• 1. Product Rule (Logarithm of a Product): \(\log_a(xy) = \log_a(x) + \log_a(y)\)

  The log of a product equals the sum of the logs

  Example: \(\log(6) = \log(2 \times 3) = \log(2) + \log(3)\)

• 2. Quotient Rule (Logarithm of a Quotient): \(\log_a\left(\displaystyle\frac{x}{y}\right) = \log_a(x) - \log_a(y)\)

  The log of a quotient equals the difference of the logs

  Example: \(\log\left(\displaystyle\frac{100}{10}\right) = \log(100) - \log(10) = 2 - 1 = 1\)

• 3. Power Rule (Logarithm of a Power): \(\log_a(x^n) = n \log_a(x)\)

  The log of a power equals the exponent times the log

  Example: \(\log(1000) = \log(10^3) = 3\log(10) = 3 \times 1 = 3\)

• 4. Change of Base Formula: \(\log_a(x) = \displaystyle\frac{\log_b(x)}{\log_b(a)}\)

  Convert between different bases (often to base 10 or \(e\) for calculator use)

  Example: \(\log_2(8) = \displaystyle\frac{\log(8)}{\log(2)} = \frac{0.903}{0.301} = 3\)

• 5. Identity Laws:

  • \(\log_a(a) = 1\) — The log of the base equals 1

  • \(\log_a(1) = 0\) — The log of 1 equals 0 (any base)

  Examples: \(\log_{10}(10) = 1\), \(\ln(e) = 1\), \(\log_5(1) = 0\)

• 6. Inverse Properties:

  • \(a^{\log_a(x)} = x\) — Exponential "undoes" logarithm

  • \(\log_a(a^x) = x\) — Logarithm "undoes" exponential

  Examples: \(10^{\log(7)} = 7\), \(\log(10^5) = 5\), \(e^{\ln(3)} = 3\)

⚠ Common Pitfalls:

• \(\log(x + y) \neq \log(x) + \log(y)\) — Logs do NOT distribute over addition!
• \(\log\left(\displaystyle\frac{x}{y}\right) \neq \frac{\log(x)}{\log(y)}\) — Use subtraction, not division!
• The argument of a logarithm must always be positive: \(\log(x)\) is only defined for \(x > 0\).
• When using change of base, both numerator and denominator must have the same new base.
• \(\displaystyle\frac{1}{\log_a(x)} = \log_x(a)\) (reciprocal), NOT \(\log_a\left(\displaystyle\frac{1}{x}\right)\)

💡 Pro Tips:

• Always check that the argument is positive before applying log rules.
• When simplifying, look for opportunities to combine or expand logs strategically.
• Use change of base when your calculator doesn't have the required base.
• Remember: \(\ln(e^x) = x\) and \(e^{\ln(x)} = x\) are your friends for natural logs!
• For multiple logs with the same base, use product/quotient rules to combine them.

Example 4: Expanding Logarithmic Expressions

Problem: Expand the following logarithmic expressions:

(a) \(\log_2(8x^3)\)

(b) \(\ln\left(\displaystyle\frac{e^2x^4}{\sqrt{y}}\right)\)

(c) \(\log\left(\displaystyle\frac{100x^2}{y^3z}\right)\)

Solution:

(a) \(\log_2(8x^3)\)

Apply product rule:

\(= \log_2(8) + \log_2(x^3)\)

Evaluate \(\log_2(8)\) and apply power rule:

Since \(8 = 2^3\), \(\log_2(8) = 3\)

\(= 3 + 3\log_2(x)\)

(b) \(\ln\left(\displaystyle\frac{e^2x^4}{\sqrt{y}}\right)\)

First rewrite \(\sqrt{y}\) as \(y^{\frac{1}{2}}\):

\(= \ln\left(\displaystyle\frac{e^2x^4}{y^{\frac{1}{2}}}\right)\)

Apply quotient rule:

\(= \ln(e^2x^4) - \ln(y^{\frac{1}{2}})\)

Apply product rule to first term:

\(= \ln(e^2) + \ln(x^4) - \ln(y^{\frac{1}{2}})\)

Apply power rule to each term:

\(= 2\ln(e) + 4\ln(x) - \frac{1}{2}\ln(y)\)

Since \(\ln(e) = 1\):

\(= 2 + 4\ln(x) - \displaystyle\frac{1}{2}\ln(y)\)

(c) \(\log\left(\displaystyle\frac{100x^2}{y^3z}\right)\)

Apply quotient rule:

\(= \log(100x^2) - \log(y^3z)\)

Apply product rule:

\(= \log(100) + \log(x^2) - [\log(y^3) + \log(z)]\)

\(= \log(100) + \log(x^2) - \log(y^3) - \log(z)\)

Apply power rule and evaluate \(\log(100) = 2\):

\(= 2 + 2\log(x) - 3\log(y) - \log(z)\)

Example 5: Condensing Logarithmic Expressions

Problem: Write each expression as a single logarithm:

(a) \(3\log(x) + 2\log(y) - \log(z)\)

(b) \(2\ln(a) - \displaystyle\frac{1}{2}\ln(b) + \ln(c)\)

Solution:

(a) \(3\log(x) + 2\log(y) - \log(z)\)

Apply power rule (reverse):

\(= \log(x^3) + \log(y^2) - \log(z)\)

Apply product rule to first two terms:

\(= \log(x^3y^2) - \log(z)\)

Apply quotient rule:

\(= \log\left(\displaystyle\frac{x^3y^2}{z}\right)\)

(b) \(2\ln(a) - \displaystyle\frac{1}{2}\ln(b) + \ln(c)\)

Apply power rule (reverse):

\(= \ln(a^2) - \ln(b^{\frac{1}{2}}) + \ln(c)\)

\(= \ln(a^2) - \ln(\sqrt{b}) + \ln(c)\)

Combine using quotient rule first:

\(= \ln\left(\displaystyle\frac{a^2}{\sqrt{b}}\right) + \ln(c)\)

Apply product rule:

\(= \ln\left(\displaystyle\frac{a^2c}{\sqrt{b}}\right)\)

Example 6: Change of Base Formula

Problem: Evaluate \(\log_5(200)\) using common logarithms (base 10). Give your answer to 3 significant figures.

Solution:

Use the change of base formula:

\(\log_5(200) = \displaystyle\frac{\log(200)}{\log(5)}\)

Using a calculator:

\(\log(200) \approx 2.301\)

\(\log(5) \approx 0.699\)

\(\log_5(200) = \displaystyle\frac{2.301}{0.699} \approx 3.292\)

Answer: 3.29 (to 3 s.f.)

Verification: \(5^{3.292} \approx 200\) ✓

Three Main Methods:

⚠ Common Pitfalls:

• When taking logs of both sides, apply the log to the entire side, not just one term.
• After taking logs, remember to use the power rule: \(\log(a^x) = x\log(a)\)
• Check solutions—some algebraic manipulations may introduce extraneous solutions.
• Be careful with signs when working with negative exponents.
• Remember: \(e^{\ln(x)} = x\) and \(\ln(e^x) = x\) (inverse properties)

💡 Strategy Guide:

1. Identify if the equation can be written with the same base
2. If yes, match bases and set exponents equal
3. If no, take logarithms of both sides (choose ln for base \(e\), log for base 10, or any base)
4. Simplify using logarithm laws
5. Isolate the variable
6. Verify your solution by substituting back

Example 7: Exponential Equations (Same Base Method)

Problem: Solve the following equations:

(a) \(3^{2x-1} = 27\)

(b) \(16^{x+1} = 4^{3x}\)

(c) \(25^x = 125^{x-1}\)

Solution:

(a) \(3^{2x-1} = 27\)

Express 27 as a power of 3:

\(27 = 3^3\)

So: \(3^{2x-1} = 3^3\)

Since bases are equal, set exponents equal:

\(2x - 1 = 3\)

\(2x = 4\)

\(x = 2\)

Check: \(3^{2(2)-1} = 3^3 = 27\) ✓

(b) \(16^{x+1} = 4^{3x}\)

Express both in terms of base 2: \(16 = 2^4\) and \(4 = 2^2\)

\((2^4)^{x+1} = (2^2)^{3x}\)

Apply power of a power rule:

\(2^{4(x+1)} = 2^{2(3x)}\)

\(2^{4x+4} = 2^{6x}\)

Set exponents equal:

\(4x + 4 = 6x\)

\(4 = 6x - 4x\)

\(4 = 2x\)

\(x = 2\)

(c) \(25^x = 125^{x-1}\)

Express both in terms of base 5: \(25 = 5^2\) and \(125 = 5^3\)

\((5^2)^x = (5^3)^{x-1}\)

\(5^{2x} = 5^{3(x-1)}\)

\(5^{2x} = 5^{3x-3}\)

Set exponents equal:

\(2x = 3x - 3\)

\(3 = 3x - 2x\)

\(x = 3\)

Example 8: Using Logarithms to Solve Exponential Equations

Problem: Solve the following equations, giving answers to 3 significant figures:

(a) \(7^x = 200\)

(b) \(5^{2x+1} = 80\)

(c) \(e^{3x-2} = 15\)

Solution:

(a) \(7^x = 200\)

Take logarithms of both sides (using common log):

\(\log(7^x) = \log(200)\)

Apply power rule:

\(x\log(7) = \log(200)\)

Solve for \(x\):

\(x = \displaystyle\frac{\log(200)}{\log(7)}\)

\(x = \displaystyle\frac{2.301}{0.845}\)

\(x \approx 2.72\)

(b) \(5^{2x+1} = 80\)

Take logarithms of both sides:

\(\log(5^{2x+1}) = \log(80)\)

Apply power rule:

\((2x+1)\log(5) = \log(80)\)

Divide both sides by \(\log(5)\):

\(2x + 1 = \displaystyle\frac{\log(80)}{\log(5)}\)

\(2x + 1 = \displaystyle\frac{1.903}{0.699}\)

\(2x + 1 = 2.723\)

\(2x = 1.723\)

\(x \approx 0.862\)

(c) \(e^{3x-2} = 15\)

Take natural logarithm of both sides (since base is \(e\)):

\(\ln(e^{3x-2}) = \ln(15)\)

Since \(\ln(e^x) = x\):

\(3x - 2 = \ln(15)\)

\(3x - 2 = 2.708\)

\(3x = 4.708\)

\(x \approx 1.57\)

Example 9: Complex Exponential Equation (IB-Style)

Problem: Solve the equation \(3^{2x} - 3^{x+1} - 18 = 0\).

Solution:

This requires substitution. Let \(u = 3^x\).

Then \(3^{2x} = (3^x)^2 = u^2\)

And \(3^{x+1} = 3 \cdot 3^x = 3u\)

Substitute into the equation:

\(u^2 - 3u - 18 = 0\)

Factor the quadratic:

\((u - 6)(u + 3) = 0\)

So \(u = 6\) or \(u = -3\)

Convert back to \(x\):

Case 1: \(u = 6\)

\(3^x = 6\)

Taking logs: \(x\log(3) = \log(6)\)

\(x = \displaystyle\frac{\log(6)}{\log(3)} = \frac{0.778}{0.477} \approx 1.63\)

Case 2: \(u = -3\)

\(3^x = -3\)

This has no real solution (exponential functions are always positive)

Answer: \(x \approx 1.63\)

Verification: \(3^{2(1.63)} - 3^{1.63+1} - 18 = 36 - 18 - 18 = 0\) ✓

Three Main Strategies:

⚠ Critical Warning - Domain Restrictions:

• ALWAYS check solutions! The argument of any logarithm must be positive.
• If \(\log(x-3) = 2\), then \(x-3 > 0\), so \(x > 3\) is required.
• Solutions that make any argument zero or negative must be rejected (extraneous solutions).
• When you exponentiate, you might introduce solutions that don't work in the original equation.
• After solving algebraically, substitute back to verify.

💡 Solving Strategy:

1. Identify domain restrictions (what makes arguments positive)
2. Use log laws to combine or simplify logarithms
3. Isolate the logarithmic expression if possible
4. Exponentiate both sides (using the appropriate base)
5. Solve the resulting equation
6. CHECK all solutions against domain restrictions

Example 10: Basic Logarithmic Equations

Problem: Solve the following equations:

(a) \(\log_3(x) = 4\)

(b) \(\log(x+5) = 2\)

(c) \(\ln(2x-1) = 3\)

Solution:

(a) \(\log_3(x) = 4\)

Convert to exponential form:

\(x = 3^4\)

\(x = 81\)

Check: \(\log_3(81) = \log_3(3^4) = 4\) ✓, and \(81 > 0\) ✓

(b) \(\log(x+5) = 2\)

This is \(\log_{10}(x+5) = 2\). Convert to exponential form:

\(x + 5 = 10^2\)

\(x + 5 = 100\)

\(x = 95\)

Check: \(x+5 = 95+5 = 100 > 0\) ✓, and \(\log(100) = 2\) ✓

(c) \(\ln(2x-1) = 3\)

Convert to exponential form (base \(e\)):

\(2x - 1 = e^3\)

\(2x = e^3 + 1\)

\(x = \displaystyle\frac{e^3 + 1}{2}\)

Using calculator: \(e^3 \approx 20.086\)

\(x = \displaystyle\frac{20.086 + 1}{2} = \frac{21.086}{2}\)

\(x \approx 10.5\)

Check: \(2(10.5) - 1 = 20 > 0\) ✓

Example 11: Combining Logarithms

Problem: Solve \(\log_2(x) + \log_2(x-3) = 2\).

Solution:

Step 1: Identify domain restrictions.

For \(\log_2(x)\): need \(x > 0\)

For \(\log_2(x-3)\): need \(x - 3 > 0\), so \(x > 3\)

Domain: \(x > 3\)

Step 2: Combine logs using product rule.

\(\log_2(x) + \log_2(x-3) = 2\)

\(\log_2(x(x-3)) = 2\)

\(\log_2(x^2 - 3x) = 2\)

Step 3: Convert to exponential form.

\(x^2 - 3x = 2^2\)

\(x^2 - 3x = 4\)

Step 4: Solve the quadratic equation.

\(x^2 - 3x - 4 = 0\)

\((x - 4)(x + 1) = 0\)

\(x = 4\) or \(x = -1\)

Step 5: Check against domain restrictions.

For \(x = 4\): \(4 > 3\) ✓ (in domain)

Check: \(\log_2(4) + \log_2(4-3) = \log_2(4) + \log_2(1) = 2 + 0 = 2\) ✓

For \(x = -1\): \(-1 \not> 3\) ✗ (NOT in domain - REJECT)

Answer: \(x = 4\)

Example 12: Logarithmic Equation with Different Terms (IB-Style)

Problem: Solve \(2\log_3(x) - \log_3(x-2) = 2\).

Solution:

Step 1: Domain restrictions.

Need \(x > 0\) and \(x - 2 > 0\), so domain: \(x > 2\)

Step 2: Apply power rule to first term.

\(2\log_3(x) = \log_3(x^2)\)

Equation becomes: \(\log_3(x^2) - \log_3(x-2) = 2\)

Step 3: Apply quotient rule.

\(\log_3\left(\displaystyle\frac{x^2}{x-2}\right) = 2\)

Step 4: Convert to exponential form.

\(\displaystyle\frac{x^2}{x-2} = 3^2\)

\(\displaystyle\frac{x^2}{x-2} = 9\)

Step 5: Solve for \(x\).

\(x^2 = 9(x-2)\)

\(x^2 = 9x - 18\)

\(x^2 - 9x + 18 = 0\)

Factor:

\((x - 6)(x - 3) = 0\)

\(x = 6\) or \(x = 3\)

Step 6: Check against domain.

For \(x = 6\): \(6 > 2\) ✓

Check: \(2\log_3(6) - \log_3(4) = \log_3(36) - \log_3(4) = \log_3\left(\frac{36}{4}\right) = \log_3(9) = 2\) ✓

For \(x = 3\): \(3 > 2\) ✓

Check: \(2\log_3(3) - \log_3(1) = 2(1) - 0 = 2\) ✓

Answer: \(x = 3\) or \(x = 6\)

Key Relationship:

\(a^x = b \iff \log_a(b) = x\)

For Exponent Questions:

• Always look for opportunities to express numbers with the same base
• Apply laws one at a time—don't skip steps
• Convert between radical and fractional exponent notation as needed
• Watch for negative exponents—they mean reciprocals, not negative values

For Logarithm Questions:

• Always check domain restrictions FIRST
• Use change of base for calculator evaluation
• Remember: logs turn multiplication into addition, division into subtraction
• After solving, always verify solutions satisfy domain requirements

Common Exam Patterns:

• Simplifying expressions using exponent/log laws
• Solving equations (same base method or taking logs)
• Converting between exponential and logarithmic forms
• Application problems involving exponential growth/decay
• Combining multiple logarithms into a single expression

Essential Functions:

• Exponents: Use the \(x^y\) or ^ button for powers
• Logarithms: Know where ln (natural log) and log (common log) buttons are
• For other bases: Use change of base: \(\log_a(x) = \displaystyle\frac{\ln(x)}{\ln(a)}\)
• Store values: Use memory (STO→) to avoid rounding errors
• Parentheses: Always use brackets for complex expressions

⚠️ Common Calculator Mistakes:

• Forgetting parentheses around fractional exponents: \(8^{(2/3)}\) not \(8^2/3\)
• Using log when you need ln (or vice versa)
• Not using change of base for bases other than 10 or \(e\)
• Rounding intermediate steps—keep full precision until the end

To master this topic, ensure you can:

• ✓ Apply all seven exponent laws correctly and efficiently
• ✓ Simplify expressions with negative and fractional exponents
• ✓ Convert between radical and exponential notation
• ✓ Apply all six logarithm laws correctly
• ✓ Expand and condense logarithmic expressions
• ✓ Use the change of base formula for calculator work
• ✓ Solve exponential equations using same base method
• ✓ Solve exponential equations using logarithms
• ✓ Solve logarithmic equations and check domain restrictions
• ✓ Convert between exponential and logarithmic forms
• ✓ Identify and reject extraneous solutions

Exponent Examples:

• Example 1: Simplifying with multiple exponent laws
• Example 2: Negative and fractional exponents
• Example 3: Algebraic expressions with variables

Logarithm Examples:

• Example 4: Expanding logarithmic expressions
• Example 5: Condensing to single logarithms
• Example 6: Change of base formula

Equation-Solving Examples:

• Example 7: Same base method (3 variations)
• Example 8: Using logarithms (3 variations)
• Example 9: Complex exponential with substitution
• Example 10: Basic logarithmic equations
• Example 11: Combining logarithms with domain checking
• Example 12: IB-style logarithmic equation

Before the Exam:

• Memorize all exponent and logarithm laws—they're not in the formula booklet!
• Practice converting between exponential and logarithmic forms instantly
• Master the change of base formula for calculator work
• Know when to use each solving method

During the Exam:

• Always show intermediate steps—marks are awarded for method
• For log equations: write "Domain: \(x > ...\)" before solving
• Check all logarithmic solutions against domain restrictions
• Use brackets liberally in calculator to avoid order of operations errors
• If stuck on same base, try taking logs instead
• Round only final answers to required precision (usually 3 s.f.)

Common Mark Losses to Avoid:

• ⚠ Not checking domain restrictions in logarithmic equations
• ⚠ Applying log laws incorrectly (especially with addition/subtraction)
• ⚠ Forgetting to show working for calculator-based problems
• ⚠ Rounding intermediate steps instead of keeping full precision
• ⚠ Missing negative solutions or incorrectly rejecting valid solutions