Hmmm.
b = a^(3/2)
d = c^(5/7)
a = b^(2/3)
c = d^(7/5)
a-c=9, so
b^(2/3) - d^(7/5) = 9
I assume we want integer solutions, so let's list the squares of cube roots and the 7th powers of 5th roots:
n n^(2/3)
1 1
8 4
27 9
64 16
125 25
216 36
n n^(7/5)
1 1
32 128
243 2187
1024 4096
I'm not getting any joy here. Any other ideas?
Logb(base a)=3/2
log d(base c)=5/7
and a-c=9
then b-d=?
Plz help me
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