[7^(2n)+16n-1],64" an "sqrt(10)(a)^(8)

If 64^(2n+1)=16^(4n-1) , what is the value of n?

Simplify the following: (i)\ (5^(n+3)-\ 6\ xx\ 5^(n+1))/(9\ xx\ 5^n-2^2\ xx\ 5^n) (ii)\ (6(8)^(n+1)+\ 16(2)^(3n-2))/(10(2)^(3n+1)-7\ (8)^n)

n in N,((1+i)/(sqrt(2)))^(8n)+((1-i)/(sqrt(2)))^(8n)=

If n in N((1+i)/(sqrt(2)))^(8n)+((1-i)/(sqrt(2)))^(8)n is

n in N((1+i)/(sqrt(2)))^(8n)+((1-i)/(sqrt(2)))^(8n)=

Use binomial theorem to show that if nge1 is an integer then , (a) 11^(n)-10n-1 is divisible by 100 (b) 3^(2n)-8n-1 is divisible by 64.

Find the values of m and n if : 4^(2m)=(root(3)(16))^(-(6)/(n))=(sqrt(8))^(2)

If 2(2n+5)=3\ (3n-10), then n= (a)5 (b) 3 (c)7 (d) 8

Simplify (6(8)^(n+1)+16(2)^(3n-2))/(10(2)^(3n+1)-7(8)^(n))