Question 790595
log(base5)y = x means {{{5^x =y}}} and {{{a*logb = logb^a}}}<P>
7*log(base5)(x-3) = log(base5)(x-3)^7 = 15<P>
{{{5^15 = (x-3)^7}}}<P>
7th root of (5^15) = x-3 = 7th root of (5^7 * 5^7 *5) = 25*7throot(5)<P>
x= 25*7throot(5) + 3