Question

(1/4)^3z-1 =16^z+2*64^z-2

Z=___

Please help

Answer 1

Z = 0.198877274

Explanation:

`((1)/(4))^(3z-1) = 16^z + 2*16^(z-2)\\4^(1-3z) = 4^(2z) + 4^{(1)/(2)}*4^(2z-4)\\4^(1-3z) = 4^(2z) + 4^{2z-4+(1)/(2)}\\4^(1-3z) = 4^(2z) + 4^{2z-(7)/(2)}\\4^(1-3z) = 4^(2z) *(1+ 4^{-(7)/(2)})\\4^(1-3z) = 4^(2z) *(1+ 2^(-7))\\4^(1-3z) = 4^(2z) *(1+ (1)/(128) )\\4^(1-3z) = 4^(2z) *((129)/(128) )\\Taking\;\; Logarithm\;\; with\;\; base\;\; 4\\Log_4(4^(1-3z)) = Log_4(4^(2z)) + Log_4((129)/(128))\\1-3z = 2z + 0.005613627712 \\5z = 0.994386372\\z = 0.198877274`

Hence, the value of Z = 0.198877274

Answer 2

The value of z is
`(3)/(8)`

Explanation:

Given equation,

`((1)/(4))^(3z-1)=16^(z+2).64^(z-2)`

`(1)/(4^(3z-1))=(4)^(2z+4).(4)^(3z-6)`

`4^(1-3z)=4^(2z+4+3z-6)`
`(a^m.a^n=a^(m+n)\text{ and }a^m=(1)/(a^(-m)))`

`4^(1-3z)=4^(5z-2)`

By comparing the exponents,

`1-3z=5z-2`

`-3z-5z=-2-1`

`-8z=-3`

`\implies z=(3)/(8)`