Question

(1/4)^3z-1 =16^z+2*64^z-2 Z=___ Please help

Answer 1

Answer:

Z = 0.198877274

Step-by-step explanation:

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

Hence, the value of Z = 0.198877274

Answer 2

Answer:

The value of z is $\frac{3}{8}$

Step-by-step explanation:

Given equation,

$(\frac{1}{4})^{3z-1}=16^{z+2}.64^{z-2}$

$\frac{1}{4^{3z-1}}=(4)^{2z+4}.(4)^{3z-6}$

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

$4^{1-3z}=4^{5z-2}$

By comparing the exponents,

$1-3z=5z-2$

$-3z-5z=-2-1$

$-8z=-3$

$\implies z=\frac{3}{8}$