Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Discover solutions to your questions from experienced professionals across multiple fields on our comprehensive Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
To solve the given equation [tex]\( 64 = 32^{x+1} \)[/tex], let's follow a step-by-step approach:
1. Rewrite the bases in terms of powers of 2:
- Notice that [tex]\( 32 \)[/tex] can be written as [tex]\( 2^5 \)[/tex].
- Also, [tex]\( 64 \)[/tex] can be written as [tex]\( 2^6 \)[/tex].
2. Express [tex]\( 32^{x+1} \)[/tex] in terms of base 2:
- Given [tex]\( 32 = 2^5 \)[/tex], it follows that [tex]\( 32^{x+1} = (2^5)^{x+1} \)[/tex].
- Using the property of exponents [tex]\((a^m)^n = a^{mn}\)[/tex], we get:
[tex]\[ (2^5)^{x+1} = 2^{5(x+1)}. \][/tex]
3. Rewrite the equation using these expressions:
- Now the equation [tex]\( 64 = 32^{x+1} \)[/tex] becomes:
[tex]\[ 2^6 = 2^{5(x+1)}. \][/tex]
4. Set the exponents equal:
- Since the bases (2) are the same, we can equate the exponents:
[tex]\[ 6 = 5(x+1). \][/tex]
5. Solve for [tex]\( x \)[/tex]:
- First, distribute the 5 on the right-hand side:
[tex]\[ 6 = 5x + 5. \][/tex]
- Next, subtract 5 from both sides to isolate the term with [tex]\( x \)[/tex]:
[tex]\[ 6 - 5 = 5x. \][/tex]
[tex]\[ 1 = 5x. \][/tex]
- Finally, divide both sides by 5 to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{1}{5}. \][/tex]
Thus, the solution is:
[tex]\[ x = 0.2. \][/tex]
1. Rewrite the bases in terms of powers of 2:
- Notice that [tex]\( 32 \)[/tex] can be written as [tex]\( 2^5 \)[/tex].
- Also, [tex]\( 64 \)[/tex] can be written as [tex]\( 2^6 \)[/tex].
2. Express [tex]\( 32^{x+1} \)[/tex] in terms of base 2:
- Given [tex]\( 32 = 2^5 \)[/tex], it follows that [tex]\( 32^{x+1} = (2^5)^{x+1} \)[/tex].
- Using the property of exponents [tex]\((a^m)^n = a^{mn}\)[/tex], we get:
[tex]\[ (2^5)^{x+1} = 2^{5(x+1)}. \][/tex]
3. Rewrite the equation using these expressions:
- Now the equation [tex]\( 64 = 32^{x+1} \)[/tex] becomes:
[tex]\[ 2^6 = 2^{5(x+1)}. \][/tex]
4. Set the exponents equal:
- Since the bases (2) are the same, we can equate the exponents:
[tex]\[ 6 = 5(x+1). \][/tex]
5. Solve for [tex]\( x \)[/tex]:
- First, distribute the 5 on the right-hand side:
[tex]\[ 6 = 5x + 5. \][/tex]
- Next, subtract 5 from both sides to isolate the term with [tex]\( x \)[/tex]:
[tex]\[ 6 - 5 = 5x. \][/tex]
[tex]\[ 1 = 5x. \][/tex]
- Finally, divide both sides by 5 to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{1}{5}. \][/tex]
Thus, the solution is:
[tex]\[ x = 0.2. \][/tex]
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.