Certainly! Let's solve the problem step by step:
We are given the equation:
[tex]\[ \log_x \left(\frac{1}{16}\right) = -4 \][/tex]
First, recall the definition of logarithms. The equation [tex]\(\log_x \left( \frac{1}{16} \right) = -4\)[/tex] means that [tex]\(x\)[/tex] raised to the power of [tex]\(-4\)[/tex] equals [tex]\(\frac{1}{16}\)[/tex]. Therefore, we can write:
[tex]\[ x^{-4} = \frac{1}{16} \][/tex]
Next, we want to solve for [tex]\(x\)[/tex].
1. Notice that [tex]\(\frac{1}{16}\)[/tex] can be rewritten as [tex]\(16^{-1}\)[/tex]:
[tex]\[ x^{-4} = 16^{-1} \][/tex]
2. Since the exponents are equal, we can set the bases equal by raising both sides to the power of [tex]\(-1\)[/tex] to eliminate the negative exponent:
[tex]\[ (x^{-4})^{-1} = (16^{-1})^{-1} \][/tex]
3. Simplifying the exponents, we get:
[tex]\[ x^4 = 16 \][/tex]
4. To find [tex]\(x\)[/tex], take the fourth root of both sides:
[tex]\[ x = \sqrt[4]{16} \][/tex]
5. We know that [tex]\(16\)[/tex] can be expressed as [tex]\(2^4\)[/tex]. Therefore:
[tex]\[ x = \sqrt[4]{2^4} \][/tex]
[tex]\[ x = 2 \][/tex]
So, the solution is:
[tex]\[ x = 2 \][/tex]
