Solving Exponential Equations - Accuplacer Aaf

How many distinct real solutions exist for the equation $3^{x^2+2x} = \frac{1}{27}$ ?
Answer: 0 First, write both sides with a base of 3. $\frac{1}{27} = 3^{-3}$ . The equation becomes $3^{x^2+2x} = 3^{-3}$ . Equating the exponents gives the quadratic equation $x^2+2x = -3$ , which can be rewritten as $x^2+2x+3=0$ . To find the number of real solutions, we check the discriminant, $b^2-4ac$ . Here, $a=1, b=2, c=3$ . The discriminant is $2^2 - 4(1)(3) = 4 - 12 = -8$ . Since the discriminant is negative, there are no real solutions to the quadratic equation.
If $7^{x-3} = 49$ , what is the value of $x^2 - x$ ?
Answer: 20 First, solve for $x$ . Rewrite 49 as $7^2$ . The equation is $7^{x-3} = 7^2$ . Equating the exponents gives $x-3 = 2$ , so $x=5$ . The question asks for the value of $x^2 - x$ . Substitute $x=5$ into the expression: $5^2 - 5 = 25 - 5 = 20$ .
For what value of $k$ is $x=3$ a solution to the equation $9^{x-1} = 3^{k}$ ?
Answer: 4 Substitute $x=3$ into the equation: $9^{3-1} = 3^k$ , which simplifies to $9^2 = 3^k$ . Since $9 = 3^2$ , we can rewrite the left side as $(3^2)^2 = 3^k$ . This simplifies to $3^4 = 3^k$ . Therefore, $k=4$ .
For which value of $x$ is $\left(\frac{1}{2}\right)^{x-3} = 8^{x+1}$ ?
Answer: $x = 0$ Rewrite both sides using base 2. We have $\frac{1}{2} = 2^{-1}$ and $8 = 2^3$ . So $\left(\frac{1}{2}\right)^{x-3} = (2^{-1})^{x-3} = 2^{-(x-3)} = 2^{-x+3}$ and $8^{x+1} = (2^3)^{x+1} = 2^{3(x+1)} = 2^{3x+3}$ . The equation becomes $2^{-x+3} = 2^{3x+3}$ . Setting exponents equal: $-x+3 = 3x+3$ , so $-x = 3x$ , which gives $-4x = 0$ , so $x = 0$ . Choice A results from solving $-x+3 = 3x-3$ (sign error). Choice B comes from $-x+3 = 3x+1$ (arithmetic error). Choice D results from solving $-x+3 = 3x+2$ (error in the constant term).
The equation $2^{3x-1} = 4^{x+2}$ has solution $x = k$ . What is the value of $3k$ ?
Answer: $3k = 15$ Rewrite with base 2: $4^{x+2} = (2^2)^{x+2} = 2^{2(x+2)} = 2^{2x+4}$ . The equation becomes $2^{3x-1} = 2^{2x+4}$ . Setting exponents equal: $3x-1 = 2x+4$ , so $x = 5$ . Therefore $3k = 3(5) = 15$ . Choice B results from solving $3x-1 = 2x+3$ (arithmetic error). Choice C comes from incorrectly getting $x = 6$ . Choice D results from solving $3x-1 = 2x+5$ (error in expanding the right side).
If $8^{2x-1} = \sqrt{2}$ , what is the value of $x$ ?
Answer: $\frac{7}{12}$ First, express both sides of the equation with a common base of 2. We have $8 = 2^3$ and $\sqrt{2} = 2^{1/2}$ . Substituting these into the equation gives $(2^3)^{2x-1} = 2^{1/2}$ . Using the power of a power rule, this becomes $2^{3(2x-1)} = 2^{1/2}$ , or $2^{6x-3} = 2^{1/2}$ . Now, set the exponents equal: $6x-3 = \frac{1}{2}$ . Add 3 to both sides: $6x = \frac{1}{2} + 3 = \frac{7}{2}$ . Finally, divide by 6: $x = \frac{7/2}{6} = \frac{7}{12}$ .
What is the solution to the equation $2^{x+4} = 4^{x+1}$ ?
Answer: 2 To solve this equation, rewrite both sides with the same base, which is 2. Since $4 = 2^2$ , the equation becomes $2^{x+4} = (2^2)^{x+1}$ . Using the power of a power rule on the right side gives $2^{x+4} = 2^{2(x+1)}$ , which simplifies to $2^{x+4} = 2^{2x+2}$ . Now that the bases are equal, the exponents must be equal: $x+4 = 2x+2$ . Subtract $x$ from both sides: $4 = x+2$ . Subtract 2 from both sides: $2 = x$ .
If $4^{x+1} = 8^{x-1}$ , what is the value of $x$ ?
Answer: 5 To solve the equation, rewrite both sides with a common base, which is 2. The equation becomes $(2^2)^{x+1} = (2^3)^{x-1}$ . Using the power of a power rule for exponents, this simplifies to $2^{2(x+1)} = 2^{3(x-1)}$ , which is $2^{2x+2} = 2^{3x-3}$ . Since the bases are equal, the exponents must be equal: $2x+2 = 3x-3$ . Solving for $x$ gives $5 = x$ .
If $5^{1-x} = \frac{1}{125}$ , what is the value of $x$ ?
Answer: 4 First, express $\frac{1}{125}$ as a power of 5. Since $125 = 5^3$ , we have $\frac{1}{125} = 5^{-3}$ . The equation is now $5^{1-x} = 5^{-3}$ . Since the bases are equal, the exponents must be equal: $1-x = -3$ . Subtracting 1 from both sides gives $-x = -4$ , so $x = 4$ .
What is the sum of all possible solutions for $x$ in the equation $10^{x^2-9} = 1$ ?
Answer: 0 For any non-zero base $b$ , the equation $b^y = 1$ implies that $y=0$ . In this problem, the exponent must be zero: $x^2 - 9 = 0$ . This is a difference of squares, which factors to $(x-3)(x+3) = 0$ . The solutions are $x=3$ and $x=-3$ . The sum of these solutions is $3 + (-3) = 0$ .
If $3^{x+2} = 9^{x-1}$ , what is the value of $x$ ?
Answer: $x = 4$ First, rewrite both sides with the same base. Since $9 = 3^2$ , we have $9^{x-1} = (3^2)^{x-1} = 3^{2(x-1)} = 3^{2x-2}$ . So the equation becomes $3^{x+2} = 3^{2x-2}$ . Since the bases are equal, the exponents must be equal: $x+2 = 2x-2$ . Solving: $x+2 = 2x-2$ , so $4 = x$ . Choice B results from incorrectly setting $x+2 = x-1$ . Choice C comes from solving $x+2 = 2x+1$ (error in expanding). Choice D results from incorrectly equating $x+2 = 2x-4$ .
If $\sqrt{5}^{x+1} = 25$ , what is the value of $x$ ?
Answer: 3 Express both sides with a common base of 5. $\sqrt{5} = 5^{1/2}$ and $25 = 5^2$ . The equation becomes $(5^{1/2})^{x+1} = 5^2$ . Simplify the left side to get $5^{\frac{1}{2}(x+1)} = 5^2$ . Set the exponents equal: $\frac{1}{2}(x+1) = 2$ . Multiply both sides by 2 to get $x+1 = 4$ . Solving for $x$ gives $x = 3$ .
The equation $27^{x} = 9^{x+2}$ has a solution of the form $x=k$ . What is the value of $2k-1$ ?
Answer: 7 First, solve for $x$ by finding a common base for 27 and 9, which is 3. $27 = 3^3$ and $9 = 3^2$ . The equation becomes $(3^3)^x = (3^2)^{x+2}$ . This simplifies to $3^{3x} = 3^{2(x+2)}$ , or $3^{3x} = 3^{2x+4}$ . Equating the exponents gives $3x = 2x+4$ . Solving for $x$ gives $x=4$ . So, $k=4$ . The question asks for the value of $2k-1$ . Substitute $k=4$ : $2(4)-1 = 8-1 = 7$ .
What is the solution to $(\frac{1}{3})^{x-2} = 81$ ?
Answer: -2 Rewrite both sides with a common base of 3. $\frac{1}{3} = 3^{-1}$ and $81 = 3^4$ . The equation becomes $(3^{-1})^{x-2} = 3^4$ . This simplifies to $3^{-1(x-2)} = 3^4$ , or $3^{-x+2} = 3^4$ . Equating the exponents gives $-x+2 = 4$ . Subtracting 2 from both sides gives $-x = 2$ , so $x = -2$ .
The equation $2^{x+3} - 2^x = 56$ has solution $x = a$ . What is the value of $2^a$ ?
Answer: $2^a = 8$ Factor the left side: $2^{x+3} - 2^x = 2^x \cdot 2^3 - 2^x = 2^x(2^3 - 1) = 2^x(8 - 1) = 7 \cdot 2^x$ . So the equation becomes $7 \cdot 2^x = 56$ . Dividing both sides by 7: $2^x = 8$ . Since $8 = 2^3$ , we have $x = 3$ . Therefore $2^a = 2^3 = 8$ . Choice A results from incorrectly getting $2^x = 4$ . Choice C comes from solving $2^x = 16$ (error in arithmetic). Choice D gives $2^x = 32$ (error in the factoring or division step).
What is the solution to the equation $3 \cdot 2^{x+1} - 5 = 43$ ?
Answer: 3 First, isolate the exponential term. Add 5 to both sides: $3 \cdot 2^{x+1} = 48$ . Then, divide by 3: $2^{x+1} = 16$ . Now, express 16 as a power of 2: $16 = 2^4$ . The equation becomes $2^{x+1} = 2^4$ . Set the exponents equal: $x+1 = 4$ . Solving for $x$ gives $x = 3$ .
What is the product of the solutions to the equation $4^{x^2} = 2^{10x-12}$ ?
Answer: 6 First, establish a common base of 2. Since $4 = 2^2$ , the equation can be rewritten as $(2^2)^{x^2} = 2^{10x-12}$ . This simplifies to $2^{2x^2} = 2^{10x-12}$ . Equating the exponents gives the quadratic equation $2x^2 = 10x-12$ . Rearrange the equation to $2x^2 - 10x + 12 = 0$ . Divide the entire equation by 2 to simplify it: $x^2 - 5x + 6 = 0$ . This factors to $(x-2)(x-3) = 0$ . The solutions are $x=2$ and $x=3$ . The product of the solutions is $2 \times 3 = 6$ .
The equation $4^x - 2^{x+3} - 48 = 0$ has a solution of the form $x = \log_2(k)$ where $k$ is a positive integer. What is the value of $k$ ?
Answer: $k = 12$ Let $y = 2^x$ . Then $4^x = (2^2)^x = (2^x)^2 = y^2$ and $2^{x+3} = 2^3 \cdot 2^x = 8y$ . The equation becomes $y^2 - 8y - 48 = 0$ . Factoring: $(y-12)(y+4) = 0$ . Since $y = 2^x > 0$ , we have $y = 12$ . So $2^x = 12$ , which means $x = \log_2(12)$ . Therefore $k = 12$ .
If $2^{x^2-5} = \frac{1}{8}$ , which of the following is a possible value of $x$ ?
Answer: $\sqrt{2}$ Rewrite $\frac{1}{8}$ as a power of 2, which is $2^{-3}$ . The equation becomes $2^{x^2-5} = 2^{-3}$ . Equating the exponents gives $x^2 - 5 = -3$ . Add 5 to both sides to get $x^2 = 2$ . Taking the square root of both sides gives $x = \pm\sqrt{2}$ . Therefore, $\sqrt{2}$ is a possible value for $x$ .
For what value of $x$ does $5^{x^2-3x} = 5^{2x-6}$ ?
Answer: $x = 2$ or $x = 3$ Since the bases are equal, we can set the exponents equal: $x^2 - 3x = 2x - 6$ . Rearranging: $x^2 - 3x - 2x + 6 = 0$ , which gives $x^2 - 5x + 6 = 0$ . Factoring: $(x-2)(x-3) = 0$ . So $x = 2$ or $x = 3$ . We should verify both solutions work in the original equation. For $x = 2$ : LHS = $5^{4-6} = 5^{-2}$ , RHS = $5^{4-6} = 5^{-2}$ ✓. For $x = 3$ : LHS = $5^{9-9} = 5^0 = 1$ , RHS = $5^{6-6} = 5^0 = 1$ ✓. Both solutions are valid. Choice A misses one solution. Choice B misses the other solution. Choice D incorrectly includes $x = -2$ (which would come from solving $x^2 - x - 6 = 0$ , a sign error).

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