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Solve the following logarithm problem for the positive solution for
x.

log_(16)x=-(1)/(2)
Answer:

Solve the following logarithm problem for the positive solution for $x$ . $\newline$ $\log _{16} x=-\frac{1}{2}$ $\newline$ Answer:

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Quotient property of logarithms

Full solution

Q. Solve the following logarithm problem for the positive solution for

x

.

\newline

\log _{16} x=-\frac{1}{2}

\newline

Answer:

Understand Equation: Understand the given logarithmic equation. $\newline$ The equation is $\log_{16} x = -\frac{1}{2}$ . This can be written as: $\newline$ $\log_{16}(x) = -\frac{1}{2}$ $\newline$ We need to find the value of $x$ that satisfies this equation.
Convert to Exponential Form: Convert the logarithmic equation to exponential form. $\newline$ Using the definition of a logarithm, we can convert the equation from logarithmic form to exponential form. The definition states that if $\log_{b}(a) = c$ , then $b^c = a$ . $\newline$ So, in our case, $16^{-(1)/(2)} = x$ .
Solve for x: Solve for x. $\newline$ We know that $16$ is $2$ raised to the power of $4$ , so we can rewrite the equation as $(2^4)^{-(1)/(2)} = x$ . $\newline$ Using the power of a power rule, which states that $(a^b)^c = a^{b*c}$ , we get $2^{4*(-(1)/(2))} = x$ . $\newline$ This simplifies to $2^{-2} = x$ .
Calculate x: Calculate the value of x. $\newline$ Since $2^{-2}$ is the same as $1/(2^2)$ , we find that $x = 1/(2^2)$ . $\newline$ Therefore, $x = 1/4$ .

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