Question

if ((9^n) * (3^2) * ((3^(-n/2))^-2)-(27)^n) / ((3^3m)*(2^3)) = 1/27 , prove that m-n = 1

Answer 1

Answer:

the whole equation is equal to 0 hence verified

Answer 2

9^n*3^2*3^n-27^n/3^3m*2^3=3^-3

:3^2n * 3^2 * 3^2 -3^3n/3^3m*2^3 = 3^-3

:3^2n+n+2 - 3^3n /3^3m*8 = 3^-3

:3^3n+2 -3^ 3n /3^3m*8 = 3^-3

taking 3^3n as a common ....

3^3n(3^2-1)/3^3m*8 =3^-3

(3^2-1)& 8 get cancelled.....

=>3^3n/3^3m=3^-3

according to law of indices...(a^x÷a^y = a^x-y)

we get..

3^3n-3m=3^-3

3 gets cancelled....

we are left with...

3n-3m=-3

=>3(n-m)=-3

=>n-m = -1

arranging :

m-n=1.....(proved)!!!

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