Question

Help,va rog frumos! :D

Answer 1

5^x+5^(-x) =2   notezi 5^x =a ,a >0 =>
=> a+ 1/a =2 => a^2 +1=2*a => a^2+1-2*a=0 => (a-1)^2=0 => a=1
dar a=5^x => 5^x=1 => x=0

Answer 2

$5^x+5^-^x=2$

$5^x+ \frac{1}{5^x}=2$

Notam  $5^x=t$

$t+ \frac{1}{t}=2$

$t^2+1-2=0$

[tex]t^2-1=0 t^2=1 t=+,- \sqrt{1} =+,-1 [/tex]

$5^x=1$

x=$log_{5}1=0$

A doua ecuatie cea cu -1 apartine multiimi vide