Question

Sa se rezolve in R. Va rog muuult chiar nu-mi iese.

Answer 1

$\it 3\cdot4^x-6^x=2\cdot9^x \Rightarrow 2\cdot9^x+6^x-3\cdot4^x=0|_{:4^x} \Rightarrow \\ \\ \\ \Rightarrow 2\cdot\left(\dfrac{9}{4}\right)^x +\left(\dfrac{6}{4}\right)^x -3=0 \Rightarrow 2\cdot\left(\dfrac{3}{2}\right)^{2x} +\left(\dfrac{3}{2}\right)^x -3=0 \Rightarrow$

$\it \Rightarrow 2\left[\cdot\left(\dfrac{3}{2}\right)^x \right]^2 +\left(\dfrac{3}{2}\right)^x -3=0 \\ \\ \\ Se\ noteaz\breve{a} \left(\dfrac{3}{2}\right)^x = t,\ t>0, iar\ ecua\c{\it t}ia\ devine:\\ \\ \\ 2t^2 + t -3= 0 \Rightarrow 2t^2-2t+3t-3=0 \Rightarrow 2t(t-1)+3(t-1)=0\Rightarrow\\ \\ \\ \Rightarrow (t-1)(2t+3) = 0 \Rightarrow\begin{cases} \it t-1=0\Rightarrow t=1\\ \\ \\ \it 2t+3=0\Rightarrow t=-\dfrac{3}{2} <0\ \ (nu\ convine)\end{cases}$

Revenim asupra notației:

$\it t=1 \Rightarrow \left(\dfrac{3}{2} \right)^x = 1 \Rightarrow x = 0$