Question

Solve: (1/4)^3z-1=16^z+2x64^z-2

Answer 1

Same thing again Apply the same rule

Answer 2

Answer:

Value of z is 3/8.

Step-by-step explanation:

Given:

$(\frac{1}{4})^{3z-1}=16^{z+2}\times64^{z-2}$

We need to solve the given expression.

Consider,

$(\frac{1}{4})^{3z-1}=16^{z+2}\times64^{z-2}$

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

Now, using law of exponent $(x^a)^b=x^{ab}$

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

$(\frac{1}{2})^{2(3z-1)}=2^{4(z+2)}\times2^{6(z-2)}$

Now using Result of Exponent on LHS $x^{-a}=(\frac{1}{x})^a$

$2^{-2(3z-1)}=2^{4(z+2)}\times2^{6(z-2)}$

Now using another law of exponent on RHS, $x^a\times a^b=x^{a+b}$

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

$2^{2-6z}=2^{4z+8+6z-12}$

$2^{2-6z}=2^{10z-4}$

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.