Question

log по основ.7(8-2х)-log8 по основ.7=log по основ.7×1/40

Answer 1

$log_7(8-2x)-log_78=log_7 frac{1}{40}\\OD3:\8-2x textgreater 0\ 2x textless 8\x textless 4\\log_7 frac{8-2x}{8}=log_7 frac{1}{40}\40*(8-2x)=8*1\8-2x=8/40\8-2x=0,2\2x=8-0,2\2x=7,8\x=7,8:2\x=3,9( textless 4)$
Ответ: 3,9