Question

6. Determinati toate solutiile reale ale ecuatiei 4 la puterea x+1 - 5 ori 2 la x + 1 = 0

Answer 1

$4^{x+1}-5* 2^{x+1}=0$
$( 2^{x+1})^2-5* 2^{x+1}=0$
 fie [tex] 2^{x+1}=t,t\ \textgreater \ 0 [/tex]
[tex] t^{2}-5t=0 [/tex]
$t=0 sau t =5$
[tex] 2^{x+1}=5 [/tex]
$x+1= log_{2}5$
$x= log_{2}5-1$
$x= log_{2} \frac{5}{2} }$