Question

ajutor dau coroana.va multumesc anticipat

Answer 1

$a^{x+y}=a^x*a^y$
$x^a*y^a=(xy)^a$
a)
$3^x*3^2-5*3^x=36$
$3^x(9-5)=36$
$3^x*4=36$
$3^x=9$
$3^x=3^2==\ \textgreater \ x=2$ ,2∈N
b)
$2^{51}+2^{52}+2^{53}=x*2^{50}$
$2^{51}=2^{50+1}=2^{50}*2$
$2^{52}=2^{50+2}=2^{50}*2^2$
$2^{53}=2^{50+3}=2^{50}*2^3$
$2^{50}*2+2^{50}*2^2+2^{50}*2^3=x*2^{50}$
$2^{50}(2+4+8)=x*2^{50}$
$2^{50}*14=x*2^{50} /:_{2^{50}}$
$14=x,x=14$, 14∈N
c)
$5^x*5^2*2^x*2+7*10^x=57000$
$10^x*50+7*10^x=57000$
$10^x(50+7)=57000$
$10^x*57=57000$
$10^x=1000$
$10^x=10^3==\ \textgreater \ x=3$ , 3∈N
d)
$12*4^x-5*4^x=7*2^{12}$
$4^x(12-5)=7*2^{12}$
$4^x*7=7*2^{12}/:_{7}$
[tex]4^x=2^{12} [/tex]
[tex] 4^x=(2^2)^x =2^{2x} [/tex]
[tex] 2^{2x}=2^{12}==\ \textgreater \ x=12[/tex], 12∈N
e)
[tex]4^{x+2}+4^{x+1}+4^x=336 [/tex]
[tex]4^x(16+4+1)=336 [/tex]
[tex]4^x*21=336 [/tex]
[tex]4^x=16 [/tex]
$4^x=4^2==\ \textgreater \ x=2$ , 2∈N
f)
$9^{43}-9^{42}-9^{41}=71*3^x$
$9^{41}(81-9-1)=71*3^x$
$9^{41}*71=71*3^x /:_{71}$
$9^{41}=3^x$
$9^{41}=(3^2)^{41}=3^{82}$
$3^{82}=3^x==>x=82$ ,82∈N
g)
$7^{x+3}-7^{x+2}-7^{x+1}=41*7^{2x}$
$7^x(7^3-7^2-7)=41*7^{2*x}$
$7^x(343-49-7)=41*7^{2*x}$
$7^x(343-49-7)=41*7^{2*x}$
$7^x*287=41*7^{2*x}:_{41}$
$7^x*7=7^{2*x}$
$7^{x+1}=7^{2*x}$
$x+1=2x$
$x=1$ ,1∈N
h)
$3^{x+3}*4^{x+1}-5*3^x*4^x=1236$
$3^x*27*4^x*4-5*12^x=1236$
$12^x*108-5*12^x=1236$
$12^x(108-5)=1236$
$12^x*103=1236$
$2^x=12==>x=1$, 1∈N