Let's solve the given equations step-by-step.
### First Equation: [tex]\(\sqrt[x]{2}=2^x\)[/tex]
The given equation is [tex]\(\sqrt[x]{2} = 2^x\)[/tex]. We can rewrite [tex]\(\sqrt[x]{2}\)[/tex] as [tex]\(2^{1/x}\)[/tex]. So, the equation becomes:
[tex]\[ 2^{1/x} = 2^x \][/tex]
Since the bases are the same, we can equate the exponents:
[tex]\[ \frac{1}{x} = x \][/tex]
To solve for [tex]\(x\)[/tex], we multiply both sides by [tex]\(x\)[/tex]:
[tex]\[ 1 = x^2 \][/tex]
Now solve for [tex]\(x\)[/tex]:
[tex]\[ x^2 = 1 \][/tex]
[tex]\[ x = \pm 1 \][/tex]
Thus, the solutions for the first equation are:
[tex]\[ x = -1 \quad \text{and} \quad x = 1 \][/tex]
### Second Equation: [tex]\(2^{x-1}=4^5\)[/tex]
The given equation is [tex]\(2^{x-1} = 4^5\)[/tex]. We know that [tex]\(4\)[/tex] can be written as [tex]\(2^2\)[/tex], so the equation becomes:
[tex]\[ 2^{x-1} = (2^2)^5 \][/tex]
Simplify the right-hand side:
[tex]\[ 2^{x-1} = 2^{10} \][/tex]
Since the bases are the same, we can equate the exponents:
[tex]\[ x - 1 = 10 \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ x = 10 + 1 \][/tex]
[tex]\[ x = 11 \][/tex]
Thus, the solution for the second equation is:
[tex]\[ x = 11 \][/tex]
### Summary of Solutions
- Solutions for [tex]\(\sqrt[x]{2}=2^x\)[/tex] are [tex]\(x = -1\)[/tex] and [tex]\(x = 1\)[/tex].
- Solution for [tex]\(2^{x-1}=4^5\)[/tex] is [tex]\(x = 11\)[/tex].
These are the complete solutions for the given equations.
