## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\text{Exact: } \dfrac{ \ln 7}{\ln 0.6 + \ln 7 }$ $\text{Approximately: } 1.356$
Since $\dfrac{3}{5} = 0.6$, the equation can be written as: $$(0.6)^x=7^{1-x}$$ Take the natural logarithm of both sides: $\ln 0.6^{x} = \ln 7^{1-x}$ WIth $\ln M^r = r \ln M$, then: $\therefore \ln 0.6^{x} = x \ln 0.6 \hspace{20pt} \text{and} \hspace{20pt} \ln 7^{1-x} = (1-x) \ln 7$ Thus, the equation above is equivalent to: $x \ln 0.6 =(1-x) \ln 7$ $x \ln 0.6 = \ln 7 -x \ln 7$ Combine the $x$ terms: $x \ln 0.6+ x \ln 7 = \ln 7$ Factor out $x$ (the common factor) then solve for $x$: $x(\ln 0.6 + \ln 7 ) = \ln 7\\\\$ $x = \boxed{\dfrac{ \ln 7}{\ln 0.6 + \ln 7 } \approx 1.356}$