0; x+2 \neq 1; \\ log_{x+2} 5=t \neq 0; log_5 (x+2)=\frac{1}{t} \neq 0;\\ 1+2t=\frac{1}{t};\\ t+2t^2=1;\\ 2t^2+t-1=0;\\ (2t-1)(t+1)=0;\\ t_1=0.5; t_2=-1; t=0.5 " alt="1+2log_{(x+2)} 5 = log_{5} (x+2);\\ x+2>0; x+2 \neq 1; \\ log_{x+2} 5=t \neq 0; log_5 (x+2)=\frac{1}{t} \neq 0;\\ 1+2t=\frac{1}{t};\\ t+2t^2=1;\\ 2t^2+t-1=0;\\ (2t-1)(t+1)=0;\\ t_1=0.5; t_2=-1; t=0.5 " align="absmiddle" class="latex-formula">
ответ: 0.5