Final: \(x \approx 0.0973\)
Step 1 & 2: Apply Log
Fill in the blank:
\(\log\left(5^{3x + 2}\right) = \log(40)\)
Check: Both sides receive identical \(\log\) operation.
Step 3: Power Rule
Bring exponent down:
\((3x + 2)\log(5) = \log(40)\)
\(3x + 2 = \frac{\log(40)}{\log(5)} \approx 2.2920\)
Divide by the constant \(\log(5)\).
Step 4: Solve for \(x\)
\(3x = \frac{\log(40)}{\log(5)} - 2\)
\(x = \frac{\log(40) - 2\log(5)}{3\log(5)}\)
Calculated Decimal: \(x \approx 0.0973\)
Final: \(x = 7\) (Exact Integer)
Stage 1: Isolate the Exponential Term
Add 10 to both sides:
\(6 \cdot 2^{x - 4} = 48\)
Divide both sides by 6:
\(2^{x - 4} = 8\)
Student work must show \(2^{x-4} = 8\).
Stage 2: Solve with Logarithms (or Inspection)
\(\log\left(2^{x - 4}\right) = \log(8)\)
\((x - 4)\log(2) = \log(8)\)
\(x - 4 = \frac{\log(8)}{\log(2)} = 3\)
\(x = 3 + 4 = 7\)
Point out: \(8 = 2^3\), so \(x - 4 = 3 \implies x = 7\)!
Frequent Misconception Alert: In Problem B, up to 30% of students attempt \(6 \cdot 2 = 12\) giving \(12^{x-4} = 48\). Stop the class immediately if this appears during the guided phase!
Teacher Resource • Answer Key
Page 3 Solutions
Problems 1–6 (General Bases)
1. \(3^{2x - 1} = 17\) Level 1
\((2x - 1)\log(3) = \log(17)\)
\(2x - 1 = \frac{\log(17)}{\log(3)} \approx 2.5789\)
\(2x = 3.5789\)
Exact: \(x = \frac{\log(17) + \log(3)}{2\log(3)} = \frac{\log(51)}{2\log(3)}\)
\(x \approx 1.7895\)
2. \(5^{x + 4} = 42\) Level 1
\((x + 4)\log(5) = \log(42)\)
\(x + 4 = \frac{\log(42)}{\log(5)} \approx 2.3223\)
\(x = 2.3223 - 4\)
Exact: \(x = \frac{\log(42)}{\log(5)} - 4\)
\(x \approx -1.6777\)
3. \(4 \cdot 2^{3x - 2} = 36\) Level 2
Divide by 4: \(2^{3x - 2} = 9\)
\((3x - 2)\log(2) = \log(9)\)
\(3x - 2 = \frac{\log(9)}{\log(2)} \approx 3.1699\)
\(3x \approx 5.1699\)
Exact: \(x = \frac{\log(9) + 2\log(2)}{3\log(2)} = \frac{\log(36)}{3\log(2)}\)
\(x \approx 1.7233\)
4. \(2 \cdot 5^{2x - 1} + 3 = 53\) Level 2
\(2 \cdot 5^{2x - 1} = 50 \implies 5^{2x - 1} = 25\)
Since \(25 = 5^2 \implies 2x - 1 = 2\)
\(2x = 3 \implies x = 1.5\)
Log Check: \(\frac{\log(25)}{\log(5)} = 2\)
\(x = 1.5\) (or \(\frac{3}{2}\))
5. Real-World Modeling Solution: Bacterial Growth Exact: t = 12 hrs
Equation: \(150 \cdot 2^{0.5t - 1} = 4800\)
Divide 150: \(2^{0.5t - 1} = 32\)
Recognize \(32 = 2^5\) (or take logs: \(0.5t - 1 = 5\))
\(0.5t = 6 \implies t = 12\)
Final Result:
\(t = 12\) hours
Doubles every 2 hrs.
6. Error Analysis Solution: Correcting the Blunder \(2 \cdot 3^{2x - 1} = 54\)
What Went Wrong:
The student incorrectly multiplied the coefficient \(2\) by the base \(3\) to get \(6\). By order of operations (PEMDAS), exponents take precedence. The coefficient must be divided out first!
Correct Algebraic Steps:
1. Divide by 2: \(3^{2x - 1} = 27\)
2. Since \(27 = 3^3 \implies 2x - 1 = 3\)
3. \(2x = 4 \implies x = 2\)
Logarithm Lockdown Teacher Key General Bases Set Verified
Teacher Resource • Answer Key
Page 4 Solutions
Problems 7–10 (Natural Logs)
7. \(e^{2x + 3} = 15\) Use \(\ln\)
\(\ln\left(e^{2x + 3}\right) = \ln(15)\)
\(2x + 3 = \ln(15) \approx 2.7081\)
\(2x = 2.7081 - 3 = -0.2919\)
Exact: \(x = \frac{\ln(15) - 3}{2}\)
\(x \approx -0.1460\)
8. \(e^{4x - 5} = 28\) Use \(\ln\)
\(\ln\left(e^{4x - 5}\right) = \ln(28)\)
\(4x - 5 = \ln(28) \approx 3.3322\)
\(4x = 8.3322\)
Exact: \(x = \frac{\ln(28) + 5}{4}\)
\(x \approx 2.0831\)
9. \(3 \cdot e^{2x + 1} = 45\) Isolate First
Divide by 3: \(e^{2x + 1} = 15\)
\(2x + 1 = \ln(15) \approx 2.7081\)
\(2x = 1.7081\)
Exact: \(x = \frac{\ln(15) - 1}{2}\)
\(x \approx 0.8540\)
10. \(5 \cdot e^{3x - 2} - 8 = 32\) Two-Step Isolate
\(5 \cdot e^{3x - 2} = 40 \implies e^{3x - 2} = 8\)
\(3x - 2 = \ln(8) \approx 2.0794\)
\(3x = 4.0794\)
Exact: \(x = \frac{\ln(8) + 2}{3}\)
\(x \approx 1.3598\)
Pedagogical Key: Emphasize Inverse Cancellation
Remind students that \(\ln(e^u) = u\) directly because \(\ln\) has base \(e\). When students write \((2x+3)\ln(e)\), emphasize that \(\ln(e) = 1\), so writing \(\ln(e)\) is redundant and simplifies immediately.
Logarithm Lockdown Teacher Key Base e Solutions Verified
Teacher Resource • Answer Key
Page 5 Solutions
Formative Check (10 Pts)
Question 1: \(3^{2x - 1} = 20\) (Use Common \(\log\)) 5 Points
Step 1 (Take \(\log\) of both sides): \(\log\left(3^{2x - 1}\right) = \log(20)\) (1 pt)
Step 2 (Apply Power Property): \((2x - 1)\log(3) = \log(20)\) (1 pt)
Step 3 (Divide by \(\log(3)\)): \(2x - 1 = \frac{\log(20)}{\log(3)} \approx \frac{1.30103}{0.47712} \approx 2.72683\) (1 pt)
Step 4 (Add 1): \(2x \approx 3.72683\) (1 pt)
Step 5 (Divide by 2): \(x \approx 1.8634\) (1 pt)
Exact Form: \(x = \frac{\log(20) + \log(3)}{2\log(3)} = \frac{\log(60)}{2\log(3)}\) Approximate: \(x \approx 1.8634\)
Question 2: \(e^{2x + 3} = 15\) (Use Natural \(\ln\)) 5 Points
Step 1 (Take \(\ln\) of both sides): \(\ln\left(e^{2x + 3}\right) = \ln(15)\) (1 pt)
Step 2 (Apply \(\ln(e) = 1\)): \(2x + 3 = \ln(15) \approx 2.70805\) (2 pts)
Step 3 (Subtract 3): \(2x = \ln(15) - 3 \approx -0.29195\) (1 pt)
Step 4 (Divide by 2): \(x \approx -0.1460\) (1 pt)
Exact Form: \(x = \frac{\ln(15) - 3}{2}\) Approximate: \(x \approx -0.1460\)
Exit Ticket Master Solutions • 10 Points All Solutions Double-Checked
