Question 944780

{{{X = B^Y}}}  is equivalent to   {{{Y = log(B, X)}}}

you are given {{{ r^s = A(t-p)^x }}}

compare it to {{{X = B^Y}}}

if {{{r}}} is base then {{{Y=s}}} and {{{X  =A(t-p)^x}}} and {{{Y = log(B, X)}}} will be

{{{s=log(r,A(t-p)^x))}}}

so, answer is D. {{{ s = log ( r, A(t-p)^x )}}}