Question

determinati numarul natural x,din egalitatea:

7^x+3-7^x+2-7^+1=41*7^2x

Answer 1

7^(x+3)-7^(x+2)-7^(x+1)=41·7^2x
7^x·7³-7^·7²+7^x·7=41·7^2x
7^x(7³-7²-7)=41·7^2x
7^x·(343-49-7)=41·7^2x
7^x·287=41·7^2x
Notam 7^x=t
287t=41t²
-41t²+287t=0
t(41t+287)=0
t1=0
-41t+287=0
-41t=-287        (-1)
t=287/41
t2=7
daca t1=0⇒7^x=7^0⇒ x=0
daca t2=7⇒7^x=7⇒x=1