a,b,c and d are natural numbers less than 1000 such that log (base a) b=7/5 and log (base c) d=4/3.Given that d-b=497,what is a+c?

$$(log_a)b=7/5$$ means that $$a^{7}=b^{5}$$ where a can't be 1.
The condition between c and is of no use. now you have to find 2 such numbers a and b which satisfies the above relation and are under 1000.
Now, $$a^{7}=b^{5}$$ should both be equal to some number $$x^{35}$$
Let us choose the smallest number after 1, i.e 2
$$a^{7}=b^{5}=2^{35}$$
which means a=$$2^{5}$$ and b=$$2^{7}$$ both of which are under 1000.
when you take the next number 3, the answers will be $$b=3^{7}$$ which exceeds 1000.
So, a=$$2^{5}$$ and b=$$2^{7}$$ i.e 32=a and 128=c. their sum is 160 which is the answer.