Find the value of r, if log r + log r2 +log r4 +log r8+log r16 +log r32 =63?

Answer 1

I will assume that your log r + log r2 +log r4 +log r8+log r16 +log r32 =63
and the base of the log is 10
is actually log r + log r^2 +log r^4 + .... + log r^32 =63
then logr + 2logr + 4logr + .... 32logr = 63
logr( 1 + 2 + 4 + ... + 32) = 63 , 6 terms inside the bracket
this is a geometric series, a = 1, r = 2, n = 6
sum(6) = 1(2^6 - 1)/(2-1) = 63 , or we could have just added them

63 logr = 63
logr = 1
r = 10