# solve the equation: 2^(2x-1)= 8^x

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2^(2x-1) = 8^x

Let us simplify 8^x :

8= 2*2*2= 2^3

==> 2^(2x-1) = (2^3)^X)

==> 2^(2x-1) = 2^3x

The bases are equal, then the powers are equal:

==> 2x-1= 3x

==> x=-1

To check:

2^(2x-1)= 8^x

2^(2(-1)-1) = 8^-1

2^(-3)=8^-3

1/ 2^3 = 1/8

1/8 = 1/8

The equation 2^(2x-1)= 8^x has to be solved for x.

Now, if the base is the same, the exponents can be equated. It is possible to write 8 as a power of 2. 8 = 2^3

8^x = (2^3)^x = 2^(3x)

The equation now becomes:

2^(2x - 1) = 2^(3x)

Equate the exponents and solve for x:

2x - 1 = 3x

x = -1

The equation 2^(2x-1)= 8^x has a solution x = -1

For the beggining, we'll write 8 as a power of 2:

8 = 2*2*2 = 2^3

Now, we'll raise to x:

8^x = (2^3)^x = 2^3x

The equation will become:

2^(2x-1)= 2^3x

We'll use one to one property:

2x - 1 = 3x

We'll sbtract 3x both sides:

-x-1 = 0

We'll add 1 both sides:

-x = 1

We'll multiply by -1 both sides:

**x = -1**

We'll verify the solution into equation:

2^(-2-1)= 8^-1

2^(-3) = 1/8

(2^3)^-1 = 1/8

1/2^3 = 1/8

1/8 = 1/8

So, x=-1 is the solution for the equation.

To silve 2^(2x-1) = 8^x

The LHS has a base 2 and the RHS has a base 8.

We can have acommon base and then equate the powers.

We know that 2 = 8^(1/3). So we convert the LHS to the base 8.

LHS = 2^(2x-1) =(8^(1/3)) ^(2x-1) = 8^((2x-1)/3) = RHS = 8^x , (as a^m)^n = a^(mn). Now the original equation is rewritten as:

8^((2x-1)/3) = 8^x. Nowboth sides have the same base 8and we can equate the powers on both sides.

(2x-1)/3 = x. Or

2x-1 = 3x.

2x-3x = 1.

-x = 1

x = -1

Let us check: 2^(2x-1) = 8^x . Put x = -1, then

LHS : 2^(2x-1)2^2 = (-2*1-1) = 2^-3 =1/8

RHS: 8^x= 8^(-1) = 1/8.

2^(2x-1)=8^x

2^(2x-1)=[(2)^3]x

2^(2x-1)=2^3x

if the bases are equate the powers

2x-1=3x

3x-2x=-1

1x=-1

x=-1

the value of x = -1

to check

2x-1=3x x=-1

(2*-1)-1=3*-1

-2-1=-3

-3=-3

LHS=RHS