a) 4^x + 2^x =20
b) 3^ 2x+1 * 4^2x = 36

D:  x ∈ R
4^x + 2^x = 20
2^2x + 2^x = 20
2^2 + 2^x + 2^x = 20
4 + 2^x + 2^x = 20
2^x + 2^x = 16
2^x + 2^x = 2^4
x+x = 4
2x = 4
x = 2 ∈ R => S: x ∈ {2}

Primu asa cred ca e in schimb al doilea ma bate rau de tot .

[tex]\displaystyle \mathtt{a)4^x+2^x=20 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~2^x=t\Rightarrow 4^x=t^2} \\ \\ t^2+t=20 \\ \\ \mathtt{t^2+t-20=0}\\ \\ \mathtt{a=1,~b=1,~c=-20}\\ \\ \mathtt{\Delta=b^2-4ac=1^2-4 \cdot 1 \cdot (-20)=1+80=81\ \textgreater \ 0}\\ \\ \mathtt{t_1= \frac{-b+ \sqrt{\Delta} }{2a} = \frac{-1+ \sqrt{81} }{2 \cdot 1}=\frac{-1+9}{2}= \frac{8}{2}=4}\\ \\ \mathtt{t_2=\frac{-b- \sqrt{\Delta} }{2a}=\frac{-1-\sqrt{81} }{2 \cdot 1}=\frac{-1-9}{2}=\frac{-10}{2}=-5}[/tex]

[tex]\displaystyle \mathtt{2^x=4\Rightarrow2^x=2^2 \Rightarrow x=2} \\ \\ \mathtt{2^x=-5\Rightarrow Nu~sunt~solutii~pentru~x \in R.} \\ \\ \mathtt{S=\{2\}}[/tex]

[tex]\displaystyle \mathtt{b)3^{2x+1} \cdot 4^{2x}=36} \\ \\ \mathtt{3^{2x} \cdot 3^1 \cdot 4^{2x}=36} \\ \\ \mathtt{12^{2x}\cdot 3=36} \\ \\ \mathtt{12^{2x}=12} \\ \\ \mathtt{2x=1} \\ \\ \mathtt{x= \frac{1}{2} }[/tex]