![image](https://tex.z-dn.net/?f=x%2Blg%281%2B2%5Ex%29%3Dxlg5%2Blg6%5C%5CODZ%3A+x+%5Cin+R%5C%5Clg10%5Ex%2Blg%281%2B2%5Ex%29%3Dlg5%5Ex%2Blg6%5C%5Clg%2810%5Ex%2A%281%2B2%5Ex%29%29%3Dlg%285%5Ex%2A6%29%5C%5C+10%5Ex%2A%281%2B2%5Ex%29%3D5%5Ex%2A6+%7C%3A5%5Ex%5C%5C+2%5Ex%2A%281%2B2%5Ex%29%3D6%5C%5C%282%5Ex%29%5E2%2B2%5Ex-6%3D0%5C%5C2%5Ex%3Dt%2C+t%3E0%5C%5Ct%5E2%2Bt-6%3D0%5C%5CD%3D1%2B24%3D5%5E2%5C%5Ct_1%3D%5Cfrac%7B-1%2B5%7D%7B2%7D%3D2%5C%5Ct_2%3D%5Cfrac%7B-1-5%7D%7B2%7D+%3C+0%5C%5C+%5C%5C2%5Ex%3Dt%5C%5C2%5Ex+%3D+2%5C%5Cx+%3D+1)
0\\t^2+t-6=0\\D=1+24=5^2\\t_1=\frac{-1+5}{2}=2\\t_2=\frac{-1-5}{2} < 0\\ \\2^x=t\\2^x = 2\\x = 1" alt="x+lg(1+2^x)=xlg5+lg6\\ODZ: x \in R\\lg10^x+lg(1+2^x)=lg5^x+lg6\\lg(10^x*(1+2^x))=lg(5^x*6)\\ 10^x*(1+2^x)=5^x*6 |:5^x\\ 2^x*(1+2^x)=6\\(2^x)^2+2^x-6=0\\2^x=t, t>0\\t^2+t-6=0\\D=1+24=5^2\\t_1=\frac{-1+5}{2}=2\\t_2=\frac{-1-5}{2} < 0\\ \\2^x=t\\2^x = 2\\x = 1" align="absmiddle" class="latex-formula">
Ответ: 1