Find the value of all integers a, b, c that satisfy 9 divided by a power of 8 multiplied by 10 divided by b power of 9 multiplied by 16 divided by c power of 15 equal to 2.

(9/8) Power a*(10/9) power b*(16/15) power c
=9^A*8^(-a)*10^b*9^(-b)*16^c*15^(-c)
=3^2A*2^(-3a)*2^b*5^b*3^(-2b)*2^4c*3^(-c)*5^(-c)
=2^(-3A+b+4c)*3^(2a-2b-c)*5^(b-c)
Because (9/8) a power *(10/9) b power *(16/15) c power =2
I.e.2^(-3a+b+4c)*3^(2a-2b-c)*5^(b-c)=2
Therefore, the index of 3 and 5 must be 0, and the index of 2 must be 1.
Therefore, we can obtain a ternary first - order system of equations about a, b, c.
-3A+b+4c=1,2a-2b-c=0, b-c=0
Solution: a=3, b=c=2,

3 To the 18th power, minus 9 to the 7th power.

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