Question

$$1.6=(1+ { 0.09 }^{ ) { y }^{ } }$$

Answer

$$y=log(0.6)^(1/)/log(0.09)^(1/)$$

Solution


Remove parentheses.
\[1.6=1+{0.09}^{}{y}^{}\]
Subtract \(1\) from both sides.
\[1.6-1={0.09}^{}{y}^{}\]
Simplify  \(1.6-1\)  to  \(0.6\).
\[0.6={0.09}^{}{y}^{}\]
Use Definition of Common Logarithm: \({b}^{a}=x\) if and only if \(log_b(x)=a\).
\[\log_{0.09}{0.6}={y}^{}\]
Use Change of Base Rule: \(\log_{b}{x}=\frac{\log_{a}{x}}{\log_{a}{b}}\).
\[\frac{\log{0.6}}{\log{0.09}}={y}^{}\]
Take the th root of both sides.
\[\sqrt[]{\frac{\log{0.6}}{\log{0.09}}}=y\]
Use Division Distributive Property: \({(\frac{x}{y})}^{a}=\frac{{x}^{a}}{{y}^{a}}\).
\[\frac{\sqrt[]{\log{0.6}}}{\sqrt[]{\log{0.09}}}=y\]
Switch sides.
\[y=\frac{\sqrt[]{\log{0.6}}}{\sqrt[]{\log{0.09}}}\]