## Algebra 2 Common Core

$x=0.00009$
Subtract $1$ from each side to obtain: $$\log{9}-\log{x}=5$$ Recall: (1) Product Property of Logarithms: $\log_a{b}+\log_a{c}=\log_a{bc}$. (2) Quotient Property of Logarithms: $\log_a{b}-\log_a{c}=\log_a{\frac{b}{c}}$. (3) Power Property of Logarithms: $\log_a{b^n} = n\cdot \log_a{b}$ Use the Quotient Property to obtain: \begin{align*} \log{9}-\log{x}&=5\\\\ \log{\frac{9}{x}}&=5 \end{align*} Recall: $$\log_a{b}=y \longleftrightarrow a^y=b$$ Use the definition above to obtain: \begin{align*} 10^5&=\frac{9}{x}\\\\ x\left(10^5\right)&=x\left(\frac{9}{x}\right)\\\\ (10^5)x&=9\\\\ \frac{(10^5)x}{10^5}&=\frac{9}{10^5}\\\\ x&=\frac{9}{10^5}\\\\ x&=0.00009 \end{align*}
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