$500 \cdot 5^{\frac{y}{3}}=1$ $\newline$ What is the solution of the equation? $\newline$ Round your answer, if necessary, to the nearest thousandth. $\newline$ $y \approx$

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Algebra 2

Solve exponential equations using common logarithms

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500 \cdot 5^{\frac{y}{3}}=1

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What is the solution of the equation?

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Round your answer, if necessary, to the nearest thousandth.

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y \approx

Write equation and solve for $y$ : Write down the equation and start solving for $y$ . $\newline$ We have the equation $500$ \cdot $5$ ^{\left(\frac{y}{ $3$ }\right)} = $1$ . $\newline$ To solve for $y$ , we first need to isolate the term with $y$ , which is $5$ ^{\left(\frac{y}{ $3$ }\right)}. $\newline$ Divide both sides of the equation by $500$ . $\newline$ $5$ ^{\left(\frac{y}{ $3$ }\right)} = \frac{ $1$ }{ $500$ }
Isolate term with y: Apply the logarithm to both sides of the equation to solve for the exponent. $\newline$ Taking the natural logarithm (ln) of both sides gives us: $\newline$ $\ln(5^{(\frac{y}{3})}) = \ln(\frac{1}{500})$
Apply logarithm to both sides: Use the power property of logarithms to bring down the exponent. $\newline$ The power property states that $\ln(a^b) = b \cdot \ln(a)$ . $\newline$ So we have: $\newline$ $\left(\frac{y}{3}\right) \cdot \ln(5) = \ln\left(\frac{1}{500}\right)$
Use power property of logarithms: Solve for $y$ . $\newline$ First, we need to calculate $\ln(5)$ and $\ln(\frac{1}{500})$ . $\newline$ Then we can multiply both sides of the equation by $3$ to isolate $y$ . $\newline$ $y = \frac{3 \cdot \ln(\frac{1}{500})}{\ln(5)}$
Solve for $y$ : Perform the calculations. $\newline$ Using a calculator, we find: $\newline$ $\ln(5) \approx 1.60944$ $\newline$ $\ln(1/500) \approx -6.21461$ $\newline$ Now substitute these values into the equation for $y$ : $\newline$ $y = \frac{3 \times (-6.21461)}{1.60944}$ $\newline$ $y \approx -11.552$
Perform calculations: Round the answer to the nearest thousandth. $\newline$ $y \approx -11.552$ $\newline$ Since we are asked to round to the nearest thousandth, the final answer is: $\newline$ $y \approx -11.552$