Certainly! Let's solve the equation [tex]\(16^x = \frac{1}{\sqrt{x}}\)[/tex] step by step.
1. Understand the equation:
[tex]\[16^x = \frac{1}{\sqrt{x}}\][/tex]
2. Express 16 and [tex]\(\frac{1}{\sqrt{x}}\)[/tex] in a simpler form:
We know that [tex]\(16\)[/tex] is [tex]\(2^4\)[/tex]. So,
[tex]\[ 16^x = (2^4)^x = 2^{4x}\][/tex]
Also, [tex]\(\frac{1}{\sqrt{x}}\)[/tex] can be written as [tex]\(x^{-\frac{1}{2}}\)[/tex].
Now, we have:
[tex]\[ 2^{4x} = x^{-\frac{1}{2}} \][/tex]
3. Find [tex]\(x\)[/tex] that satisfies [tex]\(2^{4x} = x^{-\frac{1}{2}}\)[/tex]:
4. Trial and Error:
One approach is to test likely values for [tex]\(x\)[/tex]:
Let's test [tex]\( x = \frac{1}{4} \)[/tex]:
- Calculate the left side:
[tex]\[ 16^{\frac{1}{4}} = (2^4)^{\frac{1}{4}} = 2^{4 \cdot \frac{1}{4}} = 2^1 = 2 \][/tex]
- Calculate the right side:
[tex]\[ \frac{1}{\sqrt{\frac{1}{4}}} = \frac{1}{\frac{1}{2}} = 2 \][/tex]
Both sides are equal when [tex]\( x = \frac{1}{4} \)[/tex].
5. Verify the solution:
Since [tex]\( x = \frac{1}{4} \)[/tex] satisfies the equation [tex]\(16^x = \frac{1}{\sqrt{x}}\)[/tex], it is a valid solution.
Therefore, the solution to the equation [tex]\(16^x = \frac{1}{\sqrt{x}}\)[/tex] is:
[tex]\[ x = \frac{1}{4} \][/tex]
