Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
To solve the equation [tex]\( \left(\frac{1}{4}\right)^{x+1} = 32 \)[/tex], we can take the following step-by-step approach:
1. Rewrite the equation in terms of exponents:
[tex]\[ \left(\frac{1}{4}\right)^{x+1} = 32 \][/tex]
2. Express the base [tex]\(\frac{1}{4}\)[/tex] as [tex]\((4)^{-1}\)[/tex]:
[tex]\[ \left(4^{-1}\right)^{x+1} = 32 \][/tex]
3. Simplify the left side using the properties of exponents [tex]\((a^m)^n = a^{m \cdot n}\)[/tex]:
[tex]\[ 4^{-(x+1)} = 32 \][/tex]
4. Rewrite 32 as a power of 2 for easier manipulation:
[tex]\[ 32 = 2^5 \][/tex]
5. Express the base 4 in terms of 2:
[tex]\[ 4 = 2^2 \][/tex]
Therefore,
[tex]\[ 4^{-(x+1)} = (2^2)^{-(x+1)} \][/tex]
Which simplifies to:
[tex]\[ 2^{-2(x+1)} = 2^5 \][/tex]
6. Since the bases are the same, set the exponents equal to each other:
[tex]\[ -2(x+1) = 5 \][/tex]
7. Solve for [tex]\(x\)[/tex]:
[tex]\[ -2(x+1) = 5 \][/tex]
[tex]\[ -2x - 2 = 5 \][/tex]
[tex]\[ -2x = 5 + 2 \][/tex]
[tex]\[ -2x = 7 \][/tex]
[tex]\[ x = -\frac{7}{2} \][/tex]
So, the solution to the equation [tex]\( \left(\frac{1}{4}\right)^{x+1} = 32 \)[/tex] is [tex]\( x = -\frac{7}{2} \)[/tex].
The correct answer is B. [tex]\( -\frac{7}{2} \)[/tex].
1. Rewrite the equation in terms of exponents:
[tex]\[ \left(\frac{1}{4}\right)^{x+1} = 32 \][/tex]
2. Express the base [tex]\(\frac{1}{4}\)[/tex] as [tex]\((4)^{-1}\)[/tex]:
[tex]\[ \left(4^{-1}\right)^{x+1} = 32 \][/tex]
3. Simplify the left side using the properties of exponents [tex]\((a^m)^n = a^{m \cdot n}\)[/tex]:
[tex]\[ 4^{-(x+1)} = 32 \][/tex]
4. Rewrite 32 as a power of 2 for easier manipulation:
[tex]\[ 32 = 2^5 \][/tex]
5. Express the base 4 in terms of 2:
[tex]\[ 4 = 2^2 \][/tex]
Therefore,
[tex]\[ 4^{-(x+1)} = (2^2)^{-(x+1)} \][/tex]
Which simplifies to:
[tex]\[ 2^{-2(x+1)} = 2^5 \][/tex]
6. Since the bases are the same, set the exponents equal to each other:
[tex]\[ -2(x+1) = 5 \][/tex]
7. Solve for [tex]\(x\)[/tex]:
[tex]\[ -2(x+1) = 5 \][/tex]
[tex]\[ -2x - 2 = 5 \][/tex]
[tex]\[ -2x = 5 + 2 \][/tex]
[tex]\[ -2x = 7 \][/tex]
[tex]\[ x = -\frac{7}{2} \][/tex]
So, the solution to the equation [tex]\( \left(\frac{1}{4}\right)^{x+1} = 32 \)[/tex] is [tex]\( x = -\frac{7}{2} \)[/tex].
The correct answer is B. [tex]\( -\frac{7}{2} \)[/tex].
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.