1.
[tex]\it \Big(\dfrac{1}{5}\Big)^x\cdot3^x=\sqrt[3]{\dfrac{27}{125}} \Rightarrow \Big(\dfrac{3}{5}\Big)^x=\dfrac{3}{5} \Rightarrow x=1[/tex]
2.
[tex]\it 4\cdot\Big(\dfrac{1}{16}\Big)^x+15\cdot\Big(\dfrac{1}{4}\Big)^x-4=0 \Rightarrow 4\cdot\Big(\dfrac{1}{4}\Big)^{2x}+15\cdot\Big(\dfrac{1}{4}\Big)^x-4=0\\ \\ \\ Not\breve am\ \Big(\dfrac{1}{4}\Big)^x=t,\ t>0,\ \ iar\ \ ecua\c{\it t}ia\ devine:\\ \\ 4t^2+15t-4=0 \Rightarrow 4t^2+16t-t-4=0 \Rightarrow 4t(t+4)-(t+4)=0 \Rightarrow \\ \\ \Rightarrow (t+4)(4t-1)=0 \Rightarrow t_1=-4<0\ (nu\ convine),\ \ t_2=\dfrac{1}{4}[/tex]
Revenim asupra notației și obținem:
[tex]\it t=\dfrac{1}{4} \Rightarrow \Big(\dfrac{1}{4}\Big)^x=\dfrac{1}{4} \Rightarrow x=1[/tex]
