Question-170054

Question Number 170054 by daus last updated on 15/May/22

Answered by greougoury555 last updated on 15/May/22

{\begin{cases} a = 9^{x} \\ b = 12^{x} \\ a + b = 16^{x} \end{cases} \Rightarrow 9^{x} + 12^{x} = 16^{x}

\Rightarrow {(\frac{9}{12})}^{x} + 1 = {(\frac{16}{12})}^{x}

\Rightarrow {(\frac{3}{4})}^{x} + 1 = {(\frac{4}{3})}^{x}; {(\frac{4}{3})}^{x} = (\frac{b}{a}) = y

\Rightarrow \frac{1}{y} + 1 = y; y^{2} - y - 1 = 0

\Rightarrow y = \frac{1 + \sqrt{5}}{2} = \frac{b}{a}