Question

Solutia sistemului:
3^2x-2^y=725
3^x-2^y/2=25

Answer 1

. $3^{2x}- 2^{y}=725,a,doua: 3^{x}- 2^{ \frac{y}{2} }=25,notam: 3^{x}=u,si, 2^{ \frac{y}{2} }=v$. descompunem prima ec. (u-v)(u+v)=725, a doua: u-v=25, inlocuim in prima si se obtine: 25(u+v)=725, sau u+v=29, adunam cu
u-v=25,⇒2u=54 sau $3^{x}$=27 deci x=3.Scazind cele doua ec.⇒2v=4,sau v=2, deci
$2^{ \frac{y}{2} }=2$ ⇒$\frac{y}{2}=1$, de unde y=2

Answer 2

[tex]\begin{cases}3^{2x}-2^y=725 \\ 3^x-\ 2^{\frac{y}{2}}=\ 25\end{cases} \\ Notam:\ 3^x=t,\ \ 2^{\frac{y}{2}}=q \\\;\\ Sistemul\ devine: \\\;\\ \begin{cases}t^2-q^2=725\ \ \ (1) \\ t\ -q\ =\ 25\ \ \ \ (2)\end{cases} \\\;\\ Din\ relatiile\ (1),\ (2) \Rightarrow \dfrac{t^2-q^2}{t-q} = \dfrac{725}{25}\Rightarrow\dfrac{(t-q)(t+q)}{t-q} = 29\Rightarrow \\\;\\ t+q = 29\ \ \ (3) \\\;\\ (2), (3) \Rightarrow \begin{cases}t-q=25 \\ t +q = 29\end{cases} \\\;\\ Rezulta:\ t=27,\ \ \ q=2[/tex]

[tex]Revenim\ asupra\ notatiei\ si \ rezulta: \\\;\\ 3^x = 27 \Rightarrow 3^x=3^3\Rightarrow x=3 \\ 2^{\frac{y}{2}}=2 \Rightarrow 2^{\frac{y}{2}}=2^1 \Rightarrow \dfrac{y}{2}=1 \Rightarrow y=2.[/tex]

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