🧮 Algebra Exponents And Logs (12 Focused Problems)
Recommended: 30–40 minutes. No calculator.
Problems
1.
Tags: Basic Exponents · Easy · source: Original (AMC-style)
What is $2^3 \cdot 2^4$?
A) $2^7$
B) $2^{12}$
C) $4^7$
D) $8^7$
E) $16^7$
Answer & Solution
Answer: A
Using the exponent rule: $2^3 \cdot 2^4 = 2^{3+4} = 2^7$.
2.
Tags: Basic Exponents · Easy · source: Original (AMC-style)
What is $\frac{3^5}{3^2}$?
A) $3^3$
B) $3^7$
C) $9^3$
D) $27^3$
E) $81^3$
Answer & Solution
Answer: A
Using the exponent rule: $\frac{3^5}{3^2} = 3^{5-2} = 3^3$.
3.
Tags: Power of Power · Easy · source: Original (AMC-style)
What is $(2^3)^2$?
A) $2^5$
B) $2^6$
C) $2^9$
D) $4^3$
E) $8^2$
Answer & Solution
Answer: B
Using the power rule: $(2^3)^2 = 2^{3 \cdot 2} = 2^6$.
4.
Tags: Basic Logarithms · Easy · source: Original (AMC-style)
What is $\log_2(8)$?
A) $2$
B) $3$
C) $4$
D) $8$
E) $16$
Answer & Solution
Answer: B
Since $2^3 = 8$, we have $\log_2(8) = 3$.
5.
Tags: Exponent Rules · Medium · source: Original (AMC-style)
What is $2^3 \cdot 3^3$?
A) $5^3$
B) $6^3$
C) $9^3$
D) $12^3$
E) $18^3$
Answer & Solution
Answer: B
We have $2^3 \cdot 3^3 = (2 \cdot 3)^3 = 6^3$.
6.
Tags: Logarithm Rules · Medium · source: Original (AMC-style)
What is $\log_3(27) + \log_3(9)$?
A) $3$
B) $4$
C) $5$
D) $6$
E) $7$
Answer & Solution
Answer: C
Using the product rule: $\log_3(27) + \log_3(9) = \log_3(27 \cdot 9) = \log_3(243) = \log_3(3^5) = 5$.
7.
Tags: Fractional Exponents · Medium · source: Original (AMC-style)
What is $8^{2/3}$?
A) $2$
B) $4$
C) $8$
D) $16$
E) $64$
Answer & Solution
Answer: B
We have $8^{2/3} = (8^{1/3})^2 = 2^2 = 4$.
8.
Tags: Logarithm Change of Base · Medium · source: Original (AMC-style)
What is $\log_8(64)$?
A) $1$
B) $2$
C) $3$
D) $4$
E) $8$
Answer & Solution
Answer: B
Since $8^2 = 64$, we have $\log_8(64) = 2$.
9.
Tags: Complex Exponents · Hard · source: Original (AMC-style)
If $2^x = 3^y = 6^z$, what is $\frac{1}{x} + \frac{1}{y}$ in terms of $z$?
A) $\frac{1}{z}$
B) $\frac{2}{z}$
C) $\frac{3}{z}$
D) $\frac{1}{2z}$
E) $\frac{1}{3z}$
Answer & Solution
Answer: A
Let $2^x = 3^y = 6^z = k$. Then $x = \log_2 k$, $y = \log_3 k$, and $z = \log_6 k$. We have $\frac{1}{x} + \frac{1}{y} = \frac{1}{\log_2 k} + \frac{1}{\log_3 k} = \log_k 2 + \log_k 3 = \log_k(2 \cdot 3) = \log_k 6 = \frac{1}{\log_6 k} = \frac{1}{z}$.
10.
Tags: Logarithmic Equations · Hard · source: Original (AMC-style)
Solve for $x$: $\log_2(x) + \log_2(x-1) = 3$.
A) $x = 2$
B) $x = 3$
C) $x = 4$
D) $x = 5$
E) $x = 8$
Answer & Solution
Answer: C
Using the product rule: $\log_2(x) + \log_2(x-1) = \log_2(x(x-1)) = 3$. So $x(x-1) = 2^3 = 8$, giving $x^2 - x - 8 = 0$. The positive solution is $x = \frac{1 + \sqrt{33}}{2} \approx 2.37$.
11.
Tags: Exponential Equations · Hard · source: Original (AMC-style)
Solve for $x$: $3^{2x+1} = 27^{x-1}$.
A) $x = 1$
B) $x = 2$
C) $x = 3$
D) $x = 4$
E) $x = 5$
Answer & Solution
Answer: D
Since $27 = 3^3$, we have $3^{2x+1} = (3^3)^{x-1} = 3^{3(x-1)}$. So $2x+1 = 3(x-1) = 3x-3$, giving $2x+1 = 3x-3$ and $x = 4$.
12.
Tags: Advanced Logarithms · Hard · source: Original (AMC-style)
If $\log_a(b) = 2$ and $\log_b(c) = 3$, what is $\log_a(c)$?
A) $5$
B) $6$
C) $8$
D) $9$
E) $12$
Answer & Solution
Answer: B
We have $\log_a(b) = 2$, so $b = a^2$. Also $\log_b(c) = 3$, so $c = b^3 = (a^2)^3 = a^6$. Therefore $\log_a(c) = \log_a(a^6) = 6$.
Answer Key
| # | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Ans | A | A | B | B | B | C | B | B | A | C | D | B |