Question

помогите пзалуйсто:D
$log_{x-1}(x^{2}-7x+41)=2$

Answer 1

$(x-1)^{2} = x^{2} -7x+41$
$x^{2} -2x+1=x^{2} -7x+41$
$x^{2} -2x+1-x^{2} +7x-41=0$
$5x-40=0$
$x=8$