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
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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
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