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Express the given expression without logs, in simplest form. Assume all variables represent positive values.

(e^(-2ln 12sqrty))
Answer:

Express the given expression without logs, in simplest form. Assume all variables represent positive values. $\newline$ $\left(e^{-2 \ln 12 \sqrt{y}}\right)$ $\newline$ Answer:

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Full solution

Q. Express the given expression without logs, in simplest form. Assume all variables represent positive values.

\newline

\left(e^{-2 \ln 12 \sqrt{y}}\right)

\newline

Answer:

Apply Power Rule: We are given the expression $e^{-2\ln(12\sqrt{y})}$ . To simplify this expression, we will use the property of logarithms that states $e^{\ln(x)} = x$ . This property allows us to remove the natural logarithm when it is the exponent of $e$ .
Apply Property of e: First, we need to apply the power rule of logarithms, which states that $\ln(a^b) = b\cdot\ln(a)$ . In our case, we have $-2\ln(12\sqrt{y})$ , which can be written as $\ln((12\sqrt{y})^{-2})$ .
Simplify Expression: Now, we can apply the property $e^{\ln(x)} = x$ to our expression. This means that $e^{\ln((12\sqrt{y})^{-2})}$ simplifies to $(12\sqrt{y})^{-2}$ .
Apply Negative Exponent Rule: Next, we need to simplify $(12\sqrt{y})^{-2}$ . This is the same as $(12y^{1/2})^{-2}$ , which can be simplified by applying the negative exponent rule, which states that $a^{-n} = 1/a^n$ .
Square Coefficient and Term: Applying the negative exponent rule, we get $\frac{1}{(12y^{1/2})^2}$ . Now we need to square both the coefficient $12$ and the term $y^{1/2}$ .
Square Coefficient and Term: Applying the negative exponent rule, we get $\frac{1}{(12y^{\frac{1}{2}})^2}$ . Now we need to square both the coefficient $12$ and the term $y^{\frac{1}{2}}$ . Squaring $12$ gives us $144$ , and squaring $y^{\frac{1}{2}}$ gives us $y^{\frac{1}{2} \times 2} = y^1 = y$ . So, our expression becomes $\frac{1}{144y}$ .

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