Exponents and Logs - IB Math AA SL: Comprehensive Notes
Welcome to this comprehensive set of notes on Exponents and Logarithms for the IB Mathematics: Analysis & Approaches SL curriculum. In these notes, you will find detailed explanations of key concepts, step-by-step examples, and practice questions. Mastering exponents and logarithms is crucial in IB Math AA SL, as these topics appear in various contexts, including exponential growth and decay, algebraic manipulation, and solving equations in diverse scenarios.
In this document, we have compiled:
- Complete theoretical explanations of exponents and logs.
- Key rules and formulas that you must know for the IB Exam.
- 30 example questions segmented into three levels of difficulty (Easy, Medium, and Hard) to ensure you get sufficient practice.
- Step-by-step solutions for each example, mirroring the structure of typical IB exam solutions.
1. Understanding Exponents
Exponents (also called powers) are a shorthand notation to represent repeated multiplication. If we have a base number a and an exponent n, we write:
an
This means multiply a by itself n times. For example, 5³ = 5 × 5 × 5 = 125.
Key Exponent Rules
Here are some fundamental exponent rules that are essential in IB Math AA SL:
- Product Rule: \( a^m \times a^n = a^{m + n} \)
- Quotient Rule: \( \frac{a^m}{a^n} = a^{m - n} \) (assuming \(a \neq 0\))
- Power of a Power Rule: \( (a^m)^n = a^{mn} \)
- Zero Exponent Rule: \( a^0 = 1 \) (assuming \(a \neq 0\))
- Negative Exponent Rule: \( a^{-n} = \frac{1}{a^n} \) (assuming \(a \neq 0\))
- Fractional Exponent Rule: \( a^{\frac{m}{n}} = \sqrt[n]{a^m} \) (for real \(a \ge 0\), depending on context).
2. Introduction to Logarithms
Logarithms are the inverse operations of exponents. A logarithm answers the question: "To what power must we raise the base in order to get a certain number?"
The exponential equation
ax = b
is equivalent to the logarithmic equation
\(\log_{a} (b) = x \).
In words: "x is the exponent to which we raise a to get b."
Key Logarithm Rules
Just as exponents have rules, there are corresponding logarithm rules:
- Product Rule: \( \log(a \cdot b) = \log(a) + \log(b) \)
- Quotient Rule: \( \log\left(\frac{a}{b}\right) = \log(a) - \log(b) \)
- Power Rule: \( \log(a^n) = n \cdot \log(a) \)
- Change of Base Formula: \(\log_{a}(b) = \frac{\log_{c}(b)}{\log_{c}(a)}\), where \(c\) is any positive number (commonly 10 or e).
Logarithms are essential for solving equations where the unknown is in the exponent. The IB Math AA SL curriculum often involves manipulating equations of the form \(a^x = b\), which is solved by taking \(\log\) on both sides or using the natural log (\(\ln\)) if the base is \(e\).
3. Converting Between Exponential and Logarithmic Form
A critical skill is the ability to convert an exponential equation into a logarithmic one, and vice versa. For instance:
- If \(2^3 = 8\), then in logarithmic form: \(\log_2(8) = 3\).
- If \(\log_{5}(125) = 3\), then in exponential form: \(5^3 = 125\).
This conversion is particularly useful in solving exponential and logarithmic equations, which form a core part of the IB Math AA SL curriculum.
4. Techniques for Solving Exponential and Logarithmic Equations
4.1. Solving Exponential Equations
- Taking logarithms of both sides.
- Matching bases if possible (for example, \(2^3 = 8\), \(2^4 = 16\), etc.).
- Factoring out common exponential terms.
4.2. Solving Logarithmic Equations
- Combining logarithmic expressions using product and quotient rules.
- Exponentiating to remove the logarithm once isolated (\(\log(a)=b \Rightarrow a = 10^b\) or \( e^b \) if dealing with natural logs).
- Ensuring domain validity: The argument of any log must be positive.
Next, we will dive into 30 example questions. These examples are structured with 10 easy, 10 medium, and 10 hard questions. Each question is followed by a detailed solution, mirroring the style often expected in IB exam answer booklets.
5. Example Questions and Detailed Solutions
Easy Questions (1 – 10)
Easy
1) Simplify the expression: \( 3^2 \times 3^5 \).
Solution:
Using the Product Rule of exponents: \( a^m \times a^n = a^{m+n} \).
Here, \(3^2 \times 3^5 = 3^{2+5} = 3^7 = 2187\).
Answer: \( 3^7 = 2187 \).
Using the Product Rule of exponents: \( a^m \times a^n = a^{m+n} \).
Here, \(3^2 \times 3^5 = 3^{2+5} = 3^7 = 2187\).
Answer: \( 3^7 = 2187 \).
Easy
2) Simplify the expression: \( \frac{2^6}{2^3} \).
Solution:
Using the Quotient Rule: \( \frac{a^m}{a^n} = a^{m-n} \).
\( \frac{2^6}{2^3} = 2^{6-3} = 2^3 = 8 \).
Answer: \( 8 \).
Using the Quotient Rule: \( \frac{a^m}{a^n} = a^{m-n} \).
\( \frac{2^6}{2^3} = 2^{6-3} = 2^3 = 8 \).
Answer: \( 8 \).
Easy
3) Express using positive exponents: \( x^{-4} \).
Solution:
Negative exponent rule: \( a^{-n} = \frac{1}{a^n} \).
Hence, \( x^{-4} = \frac{1}{x^4} \).
Answer: \( \frac{1}{x^4} \).
Negative exponent rule: \( a^{-n} = \frac{1}{a^n} \).
Hence, \( x^{-4} = \frac{1}{x^4} \).
Answer: \( \frac{1}{x^4} \).
Easy
4) Write in index form: \( \log_3(81) = 4 \).
Solution:
Logarithmic to exponential form: \(\log_a(b) = c \iff a^c = b\).
Here, \( \log_3(81) = 4 \) means \( 3^4 = 81 \).
Answer: \( 3^4 = 81 \).
Logarithmic to exponential form: \(\log_a(b) = c \iff a^c = b\).
Here, \( \log_3(81) = 4 \) means \( 3^4 = 81 \).
Answer: \( 3^4 = 81 \).
Easy
5) Convert to logarithmic form: \( 5^2 = 25 \).
Solution:
Exponential to logarithmic form: \( a^c = b \iff \log_a(b) = c \).
So, \( 5^2 = 25 \) becomes \( \log_5(25) = 2 \).
Answer: \( \log_5(25) = 2 \).
Exponential to logarithmic form: \( a^c = b \iff \log_a(b) = c \).
So, \( 5^2 = 25 \) becomes \( \log_5(25) = 2 \).
Answer: \( \log_5(25) = 2 \).
Easy
6) Simplify the expression: \( (x^3)^2 \).
Solution:
Power of a Power Rule: \((a^m)^n = a^{mn}\).
Thus, \((x^3)^2 = x^{3 \times 2} = x^6\).
Answer: \( x^6 \).
Power of a Power Rule: \((a^m)^n = a^{mn}\).
Thus, \((x^3)^2 = x^{3 \times 2} = x^6\).
Answer: \( x^6 \).
Easy
7) Simplify the logarithmic expression: \( \log(10^3) \). (Assuming base 10)
Solution:
Power Rule of logs: \( \log_b(a^n) = n \log_b(a) \). Also, \(\log_{10}(10) = 1\).
\( \log(10^3) = 3 \log(10) = 3 \times 1 = 3 \).
Answer: \( 3 \).
Power Rule of logs: \( \log_b(a^n) = n \log_b(a) \). Also, \(\log_{10}(10) = 1\).
\( \log(10^3) = 3 \log(10) = 3 \times 1 = 3 \).
Answer: \( 3 \).
Easy
8) Evaluate: \( 10^{\log(7)} \). (Assuming base 10 for log)
Solution:
If \(\log\) is base 10, then \(b^{\log_b(a)} = a\).
Hence, \(10^{\log_{10}(7)} = 7\).
Answer: \( 7 \).
If \(\log\) is base 10, then \(b^{\log_b(a)} = a\).
Hence, \(10^{\log_{10}(7)} = 7\).
Answer: \( 7 \).
Easy
9) Express using a single log: \( \log(4) + \log(5) \).
Solution:
Product Rule: \(\log_b(a) + \log_b(c) = \log_b(a \times c)\).
\(\log(4) + \log(5) = \log(4 \times 5) = \log(20)\).
Answer: \( \log(20) \).
Product Rule: \(\log_b(a) + \log_b(c) = \log_b(a \times c)\).
\(\log(4) + \log(5) = \log(4 \times 5) = \log(20)\).
Answer: \( \log(20) \).
Easy
10) Solve for \(x\): \( 2^x = 32 \).
Solution:
Rewrite 32 as a power of 2: \(32 = 2^5\). Then set exponents equal if bases match:
\(2^x = 2^5 \implies x = 5\).
Answer: \( x = 5 \).
Rewrite 32 as a power of 2: \(32 = 2^5\). Then set exponents equal if bases match:
\(2^x = 2^5 \implies x = 5\).
Answer: \( x = 5 \).
Medium Questions (11 – 20)
Medium
11) Simplify the expression: \( (3x^2)(4x^3) \).
Solution:
Multiply the constants and use Product Rule of exponents for x:
\((3x^2)(4x^3) = (3 \times 4) \cdot (x^2 \times x^3) = 12 \cdot x^{2+3} = 12x^5\).
Answer: \( 12x^5 \).
Multiply the constants and use Product Rule of exponents for x:
\((3x^2)(4x^3) = (3 \times 4) \cdot (x^2 \times x^3) = 12 \cdot x^{2+3} = 12x^5\).
Answer: \( 12x^5 \).
Medium
12) Solve for \(x\): \( 3^{x-2} = 9 \).
Solution:
Note that \(9 = 3^2\). Hence: \[ 3^{x-2} = 3^2. \] Equating the exponents: \(x - 2 = 2\).
Solving for \(x\), we get \(x = 4\).
Answer: \( x = 4 \).
Note that \(9 = 3^2\). Hence: \[ 3^{x-2} = 3^2. \] Equating the exponents: \(x - 2 = 2\).
Solving for \(x\), we get \(x = 4\).
Answer: \( x = 4 \).
Medium
13) Solve for \(x\): \( 10^{2x+1} = 1000 \).
Solution:
Recognize \(1000 = 10^3\). Hence: \[ 10^{2x+1} = 10^3. \] Equating exponents: \(2x + 1 = 3\).
Then, \(2x = 2 \implies x = 1\).
Answer: \( x = 1 \).
Recognize \(1000 = 10^3\). Hence: \[ 10^{2x+1} = 10^3. \] Equating exponents: \(2x + 1 = 3\).
Then, \(2x = 2 \implies x = 1\).
Answer: \( x = 1 \).
Medium
14) Express as a single log: \( \log(x) - \log(6) \).
Solution:
Quotient Rule: \(\log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right)\).
So, \(\log(x) - \log(6) = \log\left(\frac{x}{6}\right)\).
Answer: \( \log\left(\frac{x}{6}\right) \).
Quotient Rule: \(\log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right)\).
So, \(\log(x) - \log(6) = \log\left(\frac{x}{6}\right)\).
Answer: \( \log\left(\frac{x}{6}\right) \).
Medium
15) Solve for \(x\): \( \log(x) = 2 \). (Assuming base 10)
Solution:
In base 10, \(\log_{10}(x) = 2 \implies 10^2 = x\).
Hence, \(x = 100\).
Answer: \( x = 100 \).
In base 10, \(\log_{10}(x) = 2 \implies 10^2 = x\).
Hence, \(x = 100\).
Answer: \( x = 100 \).
Medium
16) Solve for \(x\): \( \ln(x) = 3 \).
Solution:
By definition of the natural logarithm (base \(e\)), \(\ln(x) = 3\) implies \(x = e^3\).
Answer: \( x = e^3 \).
By definition of the natural logarithm (base \(e\)), \(\ln(x) = 3\) implies \(x = e^3\).
Answer: \( x = e^3 \).
Medium
17) Solve for \(x\): \( \log_{2}(x) = 5 \).
Solution:
Convert to exponential form: \(\log_{2}(x) = 5 \implies 2^5 = x\).
Hence, \(x = 32\).
Answer: \( x = 32 \).
Convert to exponential form: \(\log_{2}(x) = 5 \implies 2^5 = x\).
Hence, \(x = 32\).
Answer: \( x = 32 \).
Medium
18) Solve for \(x\): \( 2^{x+1} = 32 \).
Solution:
We know \(32 = 2^5\). Thus: \[ 2^{x+1} = 2^5 \implies x + 1 = 5 \implies x = 4. \] Answer: \( x = 4 \).
We know \(32 = 2^5\). Thus: \[ 2^{x+1} = 2^5 \implies x + 1 = 5 \implies x = 4. \] Answer: \( x = 4 \).
Medium
19) Solve for \(x\): \( 3^x = 243 \).
Solution:
Recognize \(243 = 3^5\). So, \[ 3^x = 3^5 \implies x = 5. \] Answer: \( x = 5 \).
Recognize \(243 = 3^5\). So, \[ 3^x = 3^5 \implies x = 5. \] Answer: \( x = 5 \).
Medium
20) Solve for \(x\): \( 2^x = \frac{1}{8} \).
Solution:
\(\frac{1}{8} = 8^{-1} = (2^3)^{-1} = 2^{-3}\).
So, \(2^x = 2^{-3} \implies x = -3.\)
Answer: \( x = -3 \).
\(\frac{1}{8} = 8^{-1} = (2^3)^{-1} = 2^{-3}\).
So, \(2^x = 2^{-3} \implies x = -3.\)
Answer: \( x = -3 \).
Hard Questions (21 – 30)
Hard
21) Solve for \(x\): \( 3^{2x-1} = \frac{1}{9} \).
Solution:
First, note that \(\frac{1}{9} = 9^{-1} = (3^2)^{-1} = 3^{-2}\).
Hence we have: \[ 3^{2x-1} = 3^{-2}. \] Equate exponents: \(2x - 1 = -2\).
Solve for \(x\): \(2x = -1 \implies x = -\frac{1}{2}\).
Answer: \( x = -\frac{1}{2} \).
First, note that \(\frac{1}{9} = 9^{-1} = (3^2)^{-1} = 3^{-2}\).
Hence we have: \[ 3^{2x-1} = 3^{-2}. \] Equate exponents: \(2x - 1 = -2\).
Solve for \(x\): \(2x = -1 \implies x = -\frac{1}{2}\).
Answer: \( x = -\frac{1}{2} \).
Hard
22) Solve for \(x\): \( \log_{4}(x^2 + 15) = 2 \). (Corrected from x^2+1 to x^2+15 for integer solution)
Solution:
Convert to exponential form: \(\log_{4}(x^2 + 15) = 2\) means: \[ x^2 + 15 = 4^2 = 16. \] Solve \(x^2 + 15 = 16\) which leads to \(x^2 = 1\).
Hence, \(x = \pm\sqrt{1} = \pm 1\).
Domain check: \(x^2 + 15\) must be positive. Since \(x^2 \ge 0\), \(x^2+15\) is always positive. Both solutions are valid.
Answer: \( x = \pm 1 \).
Convert to exponential form: \(\log_{4}(x^2 + 15) = 2\) means: \[ x^2 + 15 = 4^2 = 16. \] Solve \(x^2 + 15 = 16\) which leads to \(x^2 = 1\).
Hence, \(x = \pm\sqrt{1} = \pm 1\).
Domain check: \(x^2 + 15\) must be positive. Since \(x^2 \ge 0\), \(x^2+15\) is always positive. Both solutions are valid.
Answer: \( x = \pm 1 \).
Hard
23) Solve the equation: \( 2^{x+2} - 2^x = 12 \).
Solution:
Rewrite \(2^{x+2} = 2^x \cdot 2^2 = 4 \cdot 2^x\). The equation becomes \(4 \cdot 2^x - 2^x = 12\). Factor out \(2^x\): \(2^x (4 - 1) = 12 \implies 2^x \cdot 3 = 12\). Hence, \(2^x = \frac{12}{3} = 4\). Since \(4 = 2^2\), we have \(2^x = 2^2 \implies x = 2\).
Answer: \( x = 2 \).
Rewrite \(2^{x+2} = 2^x \cdot 2^2 = 4 \cdot 2^x\). The equation becomes \(4 \cdot 2^x - 2^x = 12\). Factor out \(2^x\): \(2^x (4 - 1) = 12 \implies 2^x \cdot 3 = 12\). Hence, \(2^x = \frac{12}{3} = 4\). Since \(4 = 2^2\), we have \(2^x = 2^2 \implies x = 2\).
Answer: \( x = 2 \).
Hard
24) Solve the equation: \( \log(x+1) + \log(x-1) = \log(8) \). (Assuming base 10)
Solution:
First, use the Product Rule for logs: \(\log[(x+1)(x-1)] = \log(8)\). This implies \((x+1)(x-1) = 8\). Expand the left side: \(x^2 - 1 = 8 \implies x^2 = 9 \implies x = \pm 3\). Domain check: For \(\log(x+1)\) to be defined, \(x+1 > 0 \implies x > -1\). For \(\log(x-1)\) to be defined, \(x-1 > 0 \implies x > 1\). Both conditions require \(x > 1\). Thus, \(x=3\) is a valid solution. \(x=-3\) is not valid because it violates \(x>1\).
Answer: \( x = 3 \).
First, use the Product Rule for logs: \(\log[(x+1)(x-1)] = \log(8)\). This implies \((x+1)(x-1) = 8\). Expand the left side: \(x^2 - 1 = 8 \implies x^2 = 9 \implies x = \pm 3\). Domain check: For \(\log(x+1)\) to be defined, \(x+1 > 0 \implies x > -1\). For \(\log(x-1)\) to be defined, \(x-1 > 0 \implies x > 1\). Both conditions require \(x > 1\). Thus, \(x=3\) is a valid solution. \(x=-3\) is not valid because it violates \(x>1\).
Answer: \( x = 3 \).
Hard
25) Solve for \(x\): \( 5^{2x} \cdot 5^{x-1} = 125 \).
Solution:
Using the product rule for exponents on the left side: \(5^{2x + (x-1)} = 5^{3x-1}\). Recognize \(125 = 5^3\). The equation becomes: \(5^{3x-1} = 5^3\). Equating exponents: \(3x - 1 = 3 \implies 3x = 4 \implies x = \frac{4}{3}\).
Answer: \( x = \frac{4}{3} \).
Using the product rule for exponents on the left side: \(5^{2x + (x-1)} = 5^{3x-1}\). Recognize \(125 = 5^3\). The equation becomes: \(5^{3x-1} = 5^3\). Equating exponents: \(3x - 1 = 3 \implies 3x = 4 \implies x = \frac{4}{3}\).
Answer: \( x = \frac{4}{3} \).
Hard
26) Solve the equation: \( 4^{x} = 10^{x-1} \).
Solution:
Take logarithms (e.g., natural log, \(\ln\)) on both sides: \[ \ln(4^x) = \ln(10^{x-1}) \] Apply the power rule for logs: \(x \ln(4) = (x-1)\ln(10)\). Distribute on the right: \(x \ln(4) = x \ln(10) - \ln(10)\). Group terms with \(x\): \(x \ln(10) - x \ln(4) = \ln(10)\). Factor out \(x\): \(x (\ln(10) - \ln(4)) = \ln(10)\). Using quotient rule for logs: \(x \ln\left(\frac{10}{4}\right) = \ln(10) \implies x \ln(2.5) = \ln(10)\). So, \(x = \frac{\ln(10)}{\ln(2.5)}\). (Alternatively, from \(x \log(4) - x = -1\) in previous attempt: \(x(1-\log 4) = 1 \implies x = \frac{1}{1-\log 4}\) if using base 10 logs. This is equivalent.)
Answer (exact form using natural log): \(x = \frac{\ln(10)}{\ln(2.5)}\).
(Approximate value: \(\frac{2.302585}{0.916290} \approx 2.5129\))
Take logarithms (e.g., natural log, \(\ln\)) on both sides: \[ \ln(4^x) = \ln(10^{x-1}) \] Apply the power rule for logs: \(x \ln(4) = (x-1)\ln(10)\). Distribute on the right: \(x \ln(4) = x \ln(10) - \ln(10)\). Group terms with \(x\): \(x \ln(10) - x \ln(4) = \ln(10)\). Factor out \(x\): \(x (\ln(10) - \ln(4)) = \ln(10)\). Using quotient rule for logs: \(x \ln\left(\frac{10}{4}\right) = \ln(10) \implies x \ln(2.5) = \ln(10)\). So, \(x = \frac{\ln(10)}{\ln(2.5)}\). (Alternatively, from \(x \log(4) - x = -1\) in previous attempt: \(x(1-\log 4) = 1 \implies x = \frac{1}{1-\log 4}\) if using base 10 logs. This is equivalent.)
Answer (exact form using natural log): \(x = \frac{\ln(10)}{\ln(2.5)}\).
(Approximate value: \(\frac{2.302585}{0.916290} \approx 2.5129\))
Hard
27) Solve the equation: \( \ln(x) - \ln(x-2) = \ln(3) \).
Solution:
Use the Quotient Rule: \(\ln\left(\frac{x}{x-2}\right) = \ln(3)\). Since the logarithms are equal and have the same base, their arguments must be equal: \[ \frac{x}{x-2} = 3 \] Multiply by \((x-2)\): \(x = 3(x-2) \implies x = 3x - 6\). Rearrange: \(2x = 6 \implies x = 3\). Domain check: \(\ln(x)\) requires \(x > 0\). \(\ln(x-2)\) requires \(x-2 > 0 \implies x > 2\). The solution \(x=3\) satisfies \(x>2\), so it is valid.
Answer: \( x = 3 \).
Use the Quotient Rule: \(\ln\left(\frac{x}{x-2}\right) = \ln(3)\). Since the logarithms are equal and have the same base, their arguments must be equal: \[ \frac{x}{x-2} = 3 \] Multiply by \((x-2)\): \(x = 3(x-2) \implies x = 3x - 6\). Rearrange: \(2x = 6 \implies x = 3\). Domain check: \(\ln(x)\) requires \(x > 0\). \(\ln(x-2)\) requires \(x-2 > 0 \implies x > 2\). The solution \(x=3\) satisfies \(x>2\), so it is valid.
Answer: \( x = 3 \).
Hard
28) Solve for \(x\): \( \log_{2}(x+4) - \log_{2}(x-1) = 3 \).
Solution:
Use the Quotient Rule for logarithms: \(\log_{2}\left(\frac{x+4}{x-1}\right) = 3\). Convert to exponential form: \(\frac{x+4}{x-1} = 2^3 = 8\). Multiply by \((x-1)\): \(x+4 = 8(x-1) \implies x+4 = 8x - 8\). Rearrange: \(7x = 12 \implies x = \frac{12}{7}\). Domain check: For \(\log_{2}(x+4)\), we need \(x+4 > 0 \implies x > -4\). For \(\log_{2}(x-1)\), we need \(x-1 > 0 \implies x > 1\). The solution \(x = \frac{12}{7} \approx 1.71\) satisfies \(x > 1\), so it is valid.
Answer: \( x = \frac{12}{7} \).
Use the Quotient Rule for logarithms: \(\log_{2}\left(\frac{x+4}{x-1}\right) = 3\). Convert to exponential form: \(\frac{x+4}{x-1} = 2^3 = 8\). Multiply by \((x-1)\): \(x+4 = 8(x-1) \implies x+4 = 8x - 8\). Rearrange: \(7x = 12 \implies x = \frac{12}{7}\). Domain check: For \(\log_{2}(x+4)\), we need \(x+4 > 0 \implies x > -4\). For \(\log_{2}(x-1)\), we need \(x-1 > 0 \implies x > 1\). The solution \(x = \frac{12}{7} \approx 1.71\) satisfies \(x > 1\), so it is valid.
Answer: \( x = \frac{12}{7} \).
Hard
29) Solve for \(x\): \( 2^{3x} = 5^{x+1} \).
Solution:
Take natural logarithms of both sides: \(\ln(2^{3x}) = \ln(5^{x+1})\). Apply the power rule for logs: \(3x \ln(2) = (x + 1)\ln(5)\). Distribute: \(3x \ln(2) = x\ln(5) + \ln(5)\). Group terms with \(x\): \(3x \ln(2) - x \ln(5) = \ln(5)\). Factor out \(x\): \(x(3\ln(2) - \ln(5)) = \ln(5)\). Isolate \(x\): \[ x = \frac{\ln(5)}{3\ln(2) - \ln(5)} \] This can also be written as \(x = \frac{\ln(5)}{\ln(2^3) - \ln(5)} = \frac{\ln(5)}{\ln(8) - \ln(5)} = \frac{\ln(5)}{\ln(8/5)} = \frac{\ln(5)}{\ln(1.6)}\).
Answer (exact): \(x = \frac{\ln(5)}{\ln(1.6)}\).
Take natural logarithms of both sides: \(\ln(2^{3x}) = \ln(5^{x+1})\). Apply the power rule for logs: \(3x \ln(2) = (x + 1)\ln(5)\). Distribute: \(3x \ln(2) = x\ln(5) + \ln(5)\). Group terms with \(x\): \(3x \ln(2) - x \ln(5) = \ln(5)\). Factor out \(x\): \(x(3\ln(2) - \ln(5)) = \ln(5)\). Isolate \(x\): \[ x = \frac{\ln(5)}{3\ln(2) - \ln(5)} \] This can also be written as \(x = \frac{\ln(5)}{\ln(2^3) - \ln(5)} = \frac{\ln(5)}{\ln(8) - \ln(5)} = \frac{\ln(5)}{\ln(8/5)} = \frac{\ln(5)}{\ln(1.6)}\).
Answer (exact): \(x = \frac{\ln(5)}{\ln(1.6)}\).
Hard
30) Solve the equation: \( e^{2x} - 5e^x + 6 = 0 \).
Solution:
This equation is quadratic in form. Let \(y = e^x\). Then \(y^2 = (e^x)^2 = e^{2x}\). The equation becomes: \(y^2 - 5y + 6 = 0\). Factor the quadratic: \((y-2)(y-3) = 0\). So, \(y=2\) or \(y=3\). Substitute back \(y = e^x\): Case 1: \(e^x = 2 \implies x = \ln(2)\). Case 2: \(e^x = 3 \implies x = \ln(3)\). Both solutions are valid as \(e^x\) is always positive.
Answer: \( x = \ln(2) \) or \( x = \ln(3) \).
This equation is quadratic in form. Let \(y = e^x\). Then \(y^2 = (e^x)^2 = e^{2x}\). The equation becomes: \(y^2 - 5y + 6 = 0\). Factor the quadratic: \((y-2)(y-3) = 0\). So, \(y=2\) or \(y=3\). Substitute back \(y = e^x\): Case 1: \(e^x = 2 \implies x = \ln(2)\). Case 2: \(e^x = 3 \implies x = \ln(3)\). Both solutions are valid as \(e^x\) is always positive.
Answer: \( x = \ln(2) \) or \( x = \ln(3) \).
6. IB Exam Style Tips and Summary
- Show clear working: In IB Math AA SL, partial credit is often awarded. Clearly outline each step, especially when manipulating exponents and logarithms.
- Check domains: Always check domain restrictions for logarithmic functions (arguments must be positive). This is a common point in IB marking schemes.
- Simplify effectively: Look for ways to match or rewrite bases (e.g., \( 8 = 2^3 \), \( 27 = 3^3 \), \( 125 = 5^3 \), etc.) before taking logarithms.
- Memorize core log/exponent properties: This speeds up problem-solving and helps in more advanced IB Math AA SL topics like calculus and binomial expansions involving exponents.
- Practice with different bases: Don’t rely solely on base 10 or base e. Practice rewriting and converting between bases, as IB exam questions often test this skill using the change of base formula.
By diligently practicing the 30 questions provided — from the simple rules of exponents to more complex equations involving logarithms and domain constraints — you will strengthen your skills for the IB Math AA SL exam. Remember, clarity in showing each step can earn you valuable method marks, even if your final answer might have small mistakes. Also, ensure you are comfortable taking both logarithms and exponentials, as the exam can require you to switch between them seamlessly.
Finally, stay mindful of the fact that exponent and logarithm techniques often appear in other topics, including sequences and series, financial mathematics (compound interest), differentiation of exponential functions, and integration. Mastering these fundamental skills will not only help you in the algebraic manipulation sections of IB Math but also in advanced units leading up to the final exams.
Good luck with your studies and practice!
