Question

$lg(20-10^x)=5^{log_{5} (2-x)}$
Am observat ca x=1 este solutie unica, dar nu stiu cum sa redactez.

Answer 1

[tex]lg(20-10^x)=2-x\\ lg(20-10^x)=lg10^{2-x}\\ 20-10^x= \frac{10^2}{10^x}\\ Notam~10^x=t\\ 20-t= \frac{100}{t}\\ (t-10)^2=0\\ t=0\\ 10^x=10\\ x=1 [/tex]

$Obs: a^{log_ab}=b$