Answer:
Value of z is 3/8.
Step-by-step explanation:
Given:
[tex](\frac{1}{4})^{3z-1}=16^{z+2}\times64^{z-2}[/tex]
We need to solve the given expression.
Consider,
[tex](\frac{1}{4})^{3z-1}=16^{z+2}\times64^{z-2}[/tex]
[tex](\frac{1}{2^2})^{3z-1}=(2^4)^{z+2}\times(2^6)^{z-2}[/tex]
Now, using law of exponent [tex](x^a)^b=x^{ab}[/tex]
[tex]((\frac{1}{2})^2)^{3z-1}=(2)^{4(z+2)}\times(2)^{6(z-2)}[/tex]
[tex](\frac{1}{2})^{2(3z-1)}=2^{4(z+2)}\times2^{6(z-2)}[/tex]
Now using Result of Exponent on LHS [tex]x^{-a}=(\frac{1}{x})^a[/tex]
[tex]2^{-2(3z-1)}=2^{4(z+2)}\times2^{6(z-2)}[/tex]
Now using another law of exponent on RHS, [tex]x^a\times a^b=x^{a+b}[/tex]
[tex]2^{-2(3z-1)}=2^{4(z+2)+6(z-2)}[/tex]
[tex]2^{2-6z}=2^{4z+8+6z-12}[/tex]
[tex]2^{2-6z}=2^{10z-4}[/tex]
By comparing Exponent of both sides, we get
2 - 6z = 10z - 4
2 + 4 = 10z + 6z
16z = 6
z = 6/16
z = 3/8
Therefore, Value of z is 3/8.
