Exponential and Logarithmic Equations
Complete Notes & Formulae for Twelfth Grade (Precalculus)
1. Solve Exponential Equations by Rewriting the Base
One-to-One Property:
If the bases are equal, the exponents must be equal
\[ \text{If } b^S = b^T, \text{ then } S = T \]
Steps:
1. Rewrite each side as a power with the same base
2. Apply the one-to-one property (set exponents equal)
3. Solve the resulting equation for the variable
4. Check your solution in the original equation
Common Base Conversions:
| Number | As Power of 2 | As Power of 3 |
|---|---|---|
| 4 | \( 2^2 \) | — |
| 8 | \( 2^3 \) | — |
| 9 | — | \( 3^2 \) |
| 16 | \( 2^4 \) | — |
| 27 | — | \( 3^3 \) |
| 1/2 | \( 2^{-1} \) | — |
Examples:
Solve: \( 2^{x+3} = 8 \)
Rewrite 8 as \( 2^3 \):
\( 2^{x+3} = 2^3 \)
Set exponents equal: \( x + 3 = 3 \)
Solution: \( x = 0 \)
Solve: \( 9^{x-1} = 27 \)
Rewrite with base 3: \( (3^2)^{x-1} = 3^3 \)
Simplify: \( 3^{2(x-1)} = 3^3 \)
Set exponents equal: \( 2(x-1) = 3 \)
Solve: \( 2x - 2 = 3 \) → \( 2x = 5 \)
Solution: \( x = \frac{5}{2} \)
Solve: \( 4^x = \frac{1}{8} \)
Rewrite with base 2: \( (2^2)^x = 2^{-3} \)
Simplify: \( 2^{2x} = 2^{-3} \)
Set exponents equal: \( 2x = -3 \)
Solution: \( x = -\frac{3}{2} \)
2. Solve Exponential Equations Using Logarithms
Method:
Use this method when bases cannot be easily rewritten to match
Steps:
1. Isolate the exponential expression
2. Take the logarithm of both sides (usually ln or log)
3. Apply the power property: \( \log(b^x) = x\log b \)
4. Solve for the variable
5. Check your solution
Key Property:
\[ \ln(e^x) = x \quad \text{and} \quad \log(10^x) = x \]
Examples:
Solve: \( 3^x = 20 \)
Take ln of both sides: \( \ln(3^x) = \ln 20 \)
Apply power property: \( x\ln 3 = \ln 20 \)
Solve for x: \( x = \frac{\ln 20}{\ln 3} \)
Solution: \( x \approx 2.727 \)
Solve: \( 5e^{2x} = 35 \)
Divide by 5: \( e^{2x} = 7 \)
Take ln of both sides: \( \ln(e^{2x}) = \ln 7 \)
Simplify: \( 2x = \ln 7 \)
Solve: \( x = \frac{\ln 7}{2} \)
Solution: \( x \approx 0.973 \)
Solve: \( 2^{x+1} = 3^x \)
Take ln of both sides: \( \ln(2^{x+1}) = \ln(3^x) \)
Apply power property: \( (x+1)\ln 2 = x\ln 3 \)
Expand: \( x\ln 2 + \ln 2 = x\ln 3 \)
Collect x terms: \( \ln 2 = x\ln 3 - x\ln 2 \)
Factor: \( \ln 2 = x(\ln 3 - \ln 2) \)
Solution: \( x = \frac{\ln 2}{\ln 3 - \ln 2} \approx 1.710 \)
3. Solve Logarithmic Equations with One Logarithm
Method:
If \( \log_b(x) = c \), then \( x = b^c \)
Steps:
1. Isolate the logarithm on one side
2. Rewrite in exponential form
3. Solve for the variable
4. Check that solution makes argument positive
Examples:
Solve: \( \log_2(x) = 5 \)
Rewrite in exponential form: \( x = 2^5 \)
Solution: \( x = 32 \)
Solve: \( \ln(x - 3) = 2 \)
Rewrite in exponential form: \( x - 3 = e^2 \)
Solve: \( x = e^2 + 3 \)
Solution: \( x \approx 10.389 \)
Solve: \( 2\log_3(x) + 1 = 7 \)
Isolate log: \( 2\log_3(x) = 6 \)
Divide by 2: \( \log_3(x) = 3 \)
Rewrite: \( x = 3^3 \)
Solution: \( x = 27 \)
4. Solve Logarithmic Equations with Multiple Logarithms
One-to-One Property:
\[ \text{If } \log_b(M) = \log_b(N), \text{ then } M = N \]
Steps:
1. Use log properties to combine into one log on each side
2. Apply the one-to-one property (set arguments equal)
3. Solve the resulting equation
4. Check that all arguments are positive (reject extraneous solutions)
Examples:
Solve: \( \log(x) + \log(x-3) = 1 \)
Apply product property: \( \log(x(x-3)) = 1 \)
Rewrite (log base 10): \( x(x-3) = 10^1 \)
Expand: \( x^2 - 3x = 10 \)
Solve: \( x^2 - 3x - 10 = 0 \) → \( (x-5)(x+2) = 0 \)
Potential solutions: \( x = 5 \) or \( x = -2 \)
Check: x = -2 makes \( \log(-2) \) undefined (reject)
Solution: \( x = 5 \)
Solve: \( \log_2(x+1) - \log_2(x-2) = 3 \)
Apply quotient property: \( \log_2\left(\frac{x+1}{x-2}\right) = 3 \)
Rewrite: \( \frac{x+1}{x-2} = 2^3 = 8 \)
Cross multiply: \( x + 1 = 8(x - 2) \)
Expand: \( x + 1 = 8x - 16 \)
Solve: \( 17 = 7x \) → \( x = \frac{17}{7} \)
Solution: \( x = \frac{17}{7} \approx 2.43 \)
Solve: \( \ln(x) + \ln(x+2) = \ln(8) \)
Apply product property: \( \ln(x(x+2)) = \ln(8) \)
One-to-one property: \( x(x+2) = 8 \)
Expand: \( x^2 + 2x = 8 \)
Solve: \( x^2 + 2x - 8 = 0 \) → \( (x+4)(x-2) = 0 \)
Potential solutions: \( x = -4 \) or \( x = 2 \)
Check: x = -4 makes \( \ln(-4) \) undefined (reject)
Solution: \( x = 2 \)
5. Exponential Growth and Decay: Word Problems
Formulas:
Exponential Growth/Decay:
\[ A(t) = A_0(1 \pm r)^t \]
• \( A(t) \) = amount after time t
• \( A_0 \) = initial amount
• \( r \) = growth/decay rate (as decimal)
• \( t \) = time
• Use + for growth, − for decay
Continuous Growth/Decay:
\[ A(t) = A_0 e^{rt} \]
• \( e \approx 2.71828 \) (Euler's number)
• Use positive r for growth, negative for decay
Example Problems:
Problem: Population Growth
A city has 50,000 people and grows at 3% per year. What will the population be in 10 years?
Given: \( A_0 = 50{,}000 \), \( r = 0.03 \), \( t = 10 \)
Formula: \( A(t) = 50{,}000(1.03)^{10} \)
Calculate: \( A(10) = 50{,}000(1.344) \)
Answer: Approximately 67,200 people
Problem: Radioactive Decay
A substance has a half-life of 20 years. How much of a 100g sample remains after 50 years?
Half-life formula: \( A(t) = A_0 \left(\frac{1}{2}\right)^{t/h} \)
Given: \( A_0 = 100 \), \( t = 50 \), \( h = 20 \)
\( A(50) = 100 \left(\frac{1}{2}\right)^{50/20} = 100 \left(\frac{1}{2}\right)^{2.5} \)
Answer: Approximately 17.68 grams
Problem: Bacteria Growth (Continuous)
Bacteria grow continuously at 2% per hour. If there are 1000 initially, how many after 8 hours?
Formula: \( A(t) = A_0 e^{rt} \)
Given: \( A_0 = 1000 \), \( r = 0.02 \), \( t = 8 \)
\( A(8) = 1000e^{0.02(8)} = 1000e^{0.16} \)
Answer: Approximately 1174 bacteria
6. Compound Interest: Word Problems
Formulas:
Compound Interest (n times per year):
\[ A = P\left(1 + \frac{r}{n}\right)^{nt} \]
• \( A \) = final amount (future value)
• \( P \) = principal (initial investment)
• \( r \) = annual interest rate (as decimal)
• \( n \) = number of times compounded per year
• \( t \) = time in years
Continuous Compound Interest:
\[ A = Pe^{rt} \]
Used when interest is compounded continuously (infinite compounding periods)
Common Compounding Frequencies:
| Frequency | n value |
|---|---|
| Annually | 1 |
| Semiannually | 2 |
| Quarterly | 4 |
| Monthly | 12 |
| Daily | 365 |
| Continuously | Use \( A = Pe^{rt} \) |
Example Problems:
Problem: Quarterly Compounding
You invest $5,000 at 6% annual interest compounded quarterly. How much after 3 years?
Given: \( P = 5000 \), \( r = 0.06 \), \( n = 4 \), \( t = 3 \)
Formula: \( A = 5000\left(1 + \frac{0.06}{4}\right)^{4(3)} \)
\( A = 5000(1.015)^{12} \)
\( A = 5000(1.1956) \)
Answer: $5,978.09
Problem: Continuous Compounding
$10,000 is invested at 5% interest compounded continuously. Find amount after 8 years.
Given: \( P = 10{,}000 \), \( r = 0.05 \), \( t = 8 \)
Formula: \( A = 10{,}000e^{0.05(8)} \)
\( A = 10{,}000e^{0.4} \)
\( A = 10{,}000(1.4918) \)
Answer: $14,918.25
Problem: Finding Time
How long for $2,000 to grow to $3,000 at 4% compounded monthly?
Given: \( P = 2000 \), \( A = 3000 \), \( r = 0.04 \), \( n = 12 \)
Equation: \( 3000 = 2000\left(1 + \frac{0.04}{12}\right)^{12t} \)
Simplify: \( 1.5 = (1.00333)^{12t} \)
Take ln: \( \ln(1.5) = 12t \ln(1.00333) \)
Solve: \( t = \frac{\ln(1.5)}{12\ln(1.00333)} \)
Answer: Approximately 10.16 years
7. Quick Reference Summary
Key Methods:
Exponential: Rewrite bases or take logs
Logarithmic: Convert to exponential or use properties
Growth/Decay: \( A = A_0(1 \pm r)^t \) or \( A = A_0e^{rt} \)
Compound Interest: \( A = P\left(1 + \frac{r}{n}\right)^{nt} \) or \( A = Pe^{rt} \)
Always check for extraneous solutions in log equations!
📚 Study Tips
✓ For exponentials: try common base first, then use logarithms
✓ For logarithms: combine using properties, then convert to exponential
✓ Always check that log arguments are positive
✓ Growth rate as decimal: 5% = 0.05
✓ For continuous compounding, use e; otherwise use the formula with n