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В силу формулы $\log_ab\cdot\log_cd=\log_ad\cdot\log_cd$ , имеем:

$\log_3(x-1)\cdot\log_{x-1}(3^x+1)=\log_3(3^x+1)\cdot\log_{x-1}(x-1)=\log_3(3^x+1)$ , при условиях $x>1,x\ne2$ .

Используя формулу $\log_a(bc)=\log_ab+\log_ac$ , получим:

$\log_3(3^{x+1}+3)=\log_3(3(3^x+1))=\log_33+\log_3(3^x+1)=1+\log_3(3^x+1)</sc</article></body></html>$

Задание №4162 ЕГЭ по Математике