Question

4^x-1*(0.5)^3-2x=(1/8)^x

Answer 1

4^(x-1)  * (0.5)^(3-2x)=(1/8)^x
2^2(x-1)  *   2^(-(3-2x))   =   2^-3x
 2^(2x-2-3+2x)  = 2^(-3x)
2^(4x-5)=2^(-3x)
4x-5=-3x
7x=5
x=5/7

Answer 2

Answer:

The solution of the expression is $x=\frac{5}{7}$              

Step-by-step explanation:

Given : Expression $4^{x-1}\times (0.5)^{3-2x}=(\frac{1}{8})^x$

To find : Solve the expression ?

Solution :

$4^{x-1}\times (0.5)^{3-2x}=(\frac{1}{8})^x$

We solve the expression,

$(2^2)^{x-1}\times (\frac{1}{2})^{3-2x}=(\frac{1}{2^3})^x$

$(2)^{2x-2}\times (2^{-1})^{3-2x}=2^{-3x}$

$(2)^{2x-2}\times 2^{2x-3}=2^{-3x}$

$(2)^{2x-2+2x-3}=2^{-3x}$

$(2)^{4x-5}=2^{-3x}$

$(2)^{4x}\times 2^{-5}=2^{-3x}$

$\frac{(2)^{4x}}{2^{5}}=2^{-3x}$

$\frac{(2)^{4x}}{2^{-3x}}=2^{5}$

$(2)^{4x+3x}=2^{5}$

$(2)^{7x}=2^{5}$

On comparing the powers as base are same,

$7x=5$

$x=\frac{5}{7}$

Therefore, The solution of the expression is $x=\frac{5}{7}$