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]
