If a=3^p, b=3^q, c=3^r, and d=3^s and if p,q,r,and s are positive integers, determine the smallest value of p+q+r+s such that: a^2+b^3+c^5=d^7

Question

hartnn · Answer

interesting.

hartnn · Answer

a=3^p --> p = log[3]  a 
p+q+r+s = log[3]  abcd

anonymous · Answer

log?

anonymous · Answer

i kind of understand the first part but i am not quite sure about the second part p+q+r+s = log[3]  abcd

hartnn · Answer

a^2+b^3 +c^5 = d^7 gives 
3^{2p} + 3^ {3q} + 3^ {5r} = 3^ {7s}
oh, i have used that log x + log y = log xy

hartnn · Answer

i am just trying, idk the solution

anonymous · Answer

it's ok, we should try any attempt

hartnn · Answer

but now did u get this p+q+r+s = log[3]  (abcd) ?

anonymous · Answer

yep

hartnn · Answer

is this calculus related ? which topic does this question belong ?

anonymous · Answer

no, it is not caculus related. this is an exponent question

anonymous · Answer

Well d must be odd which means s must also be odd.

anonymous · Answer

why does d must be odd?

turingtest · Answer

*

anonymous · Answer

d must be odd because |dw:1349932869824:dw|