2^(p+2)=16(4)^3/2q, log2 6-log2(15q-3p)=1?

Question

solve the simultaneous equations:
answer: p=13/2 ,q=2

Answer 1

16(4)^((3/2)q) = 4^(3q/2+2) = 2^(3q+4) = 2^(p+2) so that
3q + 4 = p + 2 or
3q + 2 = p

log2(x) = log base 2 of x
log2(6) - log2(15q-3p) =
log2(6/(15q-3p)) =
log2(2/(5q-p)) = 1 = log2(2) so that
2/(5q-p) = 2 or
5q-p = 1

Since p = 5q-1
3q + 2 = 5q - 1 or
q = 3/2
and
p = 13/2

Note this differs from the claimed answer of q=2

Answer 2

you're on the properly suited track. log2(x) + log2(5) = 6 combine them utilising the valuables of logs. log2(5x) = 6 Then translate the logarithm into an exponential equation: 2^6 = 5x sixty 4 = 5x sixty 4/5 = x