4^[x-1over2]=2x^2

Answer 1

To solve for x in the equation given, we can start by taking the natural logarithm of both sides:

ln(4^(x-1/2)) = ln(2x^2)

Using the property of logarithms that states ln(a^b) = b*ln(a), we can rewrite the left side of the equation as:

(x-1/2) * ln(4) = 2*ln(2x)

Next, we can simplify this expression by using the fact that ln(4) = ln(2^2) = 2*ln(2):

(x-1/2) * 2*ln(2) = 2*ln(2x)

Now, we can distribute the terms and simplify further:

2x*ln(2) - ln(2) = 2*ln(2x)

2x*ln(2) - ln(2) = 2*ln(2) + 2*ln(x)

2x*ln(2) - ln(2) = ln(4) + ln(x^2)

2x*ln(2) - ln(2) = ln(4x^2)

Now, we can raise both sides as e (base of natural logarithm) to eliminate the logarithms:

e^(2x*ln(2) - ln(2)) = e^(ln(4x^2))

This simplifies to:

e^(2x*ln(2)) * e^(-ln(2)) = 4x^2

Using the property that e^(ln(a)) = a , we get:

2^2x * 1/2 = 4x^2

2^x = 4x^2

Now, we can rewrite this equation as:

2^x = 2^(2x)

Since the bases are the same, we can set the exponents equal to each other:

x = 2x

This simplifies to:

x = 0

Therefore, the solution to the given equation is x = 0.