Question

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

Answer 1

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