Question

(0,25)^(x^2-4)=2^(x^2-1)

Answer 1

$( \frac{1}{4})^{x^2-4}=2^{x^2-1}$
$2^{-2(x^2-4)}=2^{x^2-1}$
$-2(x^2-4)=x^2-1$
$-2x^2+8=x^2-1$
$-3x^2=-9$
$x^2=3$
$x= \sqrt{3}$;  $x=- \sqrt{3}$

Answer 2

To solve the equation (0.25)^(x^2-4) = 2^(x^2-1), we can start by taking the logarithm of both sides of the equation.

Using the property log(a^b) = b * log(a), we can rewrite the equation as:

(x^2-4) * log(0.25) = (x^2-1) * log(2)

Now, let's solve for x by isolating the terms involving x.

(x^2-4) * log(0.25) - (x^2-1) * log(2) = 0

Simplifying the equation further:

(x^2-4) * log(0.25) = (x^2-1) * log(2)

Distributing log(0.25) and log(2):

x^2 * log(0.25) - 4 * log(0.25) = x^2 * log(2) - log(2)

Rearranging the equation:

x^2 * (log(0.25) - log(2)) = -log(2) + 4 * log(0.25)

Now, we can substitute the values of log(0.25) and log(2):

x^2 * (log(0.25) - log(2)) = -log(2) + 4 * log(0.25)

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

Simplifying further:

-2 * x^2 * log(2) = -log(2) + 8 * log(2)

Dividing both sides by -2 * log(2):

x^2 = (-log(2) + 8 * log(2)) / (-2 * log(2))

Simplifying the right side:

x^2 = (8 * log(2) - log(2)) / (2 * log(2))

x^2 = 7 * log(2) / (2 * log(2))

Canceling out the log(2) terms:

x^2 = 7 / 2

Taking the square root of both sides:

x = ±√(7/2)

Therefore, the solutions to the equation (0.25)^(x^2-4) = 2^(x^2-1) are x = √(7/2) and x = -√(7/2).