Question

Am nevoie de 10..urgent

Answer 1

[tex]lg(2^x-1)= \frac{lg2+lg(2^x+3)}{2} \\ 2lg(2^x-1)=lg2(2^x+3) \\ (2^x-1)^2=2(2^x+3) \\ (2^x)^2-2*2^x+1=2*2^x+6 \\ (2^x)^2-4*2^x-5=0 \\ fie 2^x=t,t\ \textgreater \ 0 \\ t^2-4t-5=0 \\ t_{1}=-1; t_{2}=5 \\ revenim \\ 2^x=5 \\ x= log_{2}5 [/tex]