А)logx(4)=2  (x-основание)б)log(2x+2)(16x)=1  (2x+2-основание)

[latex]a) \ \log_{x}4 = 2, \ x > 0, \ x \ne 1\\\\ \log_{x}4 = \log_{x}x^2\\\\ 4 = x^2, \ x = \pm2\\\\ \boxed{x = 2}\\\\ b) \ \log_{2x+2}16x = 1, \ 2x + 2 > 0, \  2x + 2 \ne 1\\\\ \log_{2x+2}16x = \log_{2x+2}2x+2\\\\ 16x = 2x + 2, \ 14x = 2\\\\ \boxed{x =\frac{1}{7}} [/latex]