128/32 = r^2
r=2, so y=32*2=64
Or, evaluate the logs. The sequence is thus
5,log2(y),7
so, log2(y) = 6, and y=64
log a^2b^3 = 2loga + 3logb
log a/b = loga - logb
2loga + 3logb = x
loga - logb = y
2loga + 3logb = x
2loga - 2logb = 2y
5logb = x-2y
logb = (x-2y)/5
solve for loga in like wise.
1. The sequence log2 32, log2 y, log2 128, ... forms an arithmetic sequence. What is the value of y?
2. If log a^2 b^3 = x and log (a/b) = y, what are the values of log a and log b?
1 answer