Solve for x 2^(x^2-4)=3^(x^2-4)

$2^{x^{2} - 4} = 3^{x^{2} - 4}$

Step 1

Take the log of both sides of the equation.

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

Step 2

Expand $\ln (2^{x^{2} - 4})$ by moving $x^{2} - 4$ outside the logarithm.

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

Step 3

Expand $\ln (3^{x^{2} - 4})$ by moving $x^{2} - 4$ outside the logarithm.

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

Step 4

Solve the equation for $x$ .

Simplify $(x^{2} - 4) \ln (2)$ .

Start simplifying.

$0 + 0 + (x^{2} - 4) \ln (2) = (x^{2} - 4) \ln (3)$

Simplify by adding zeros.

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

Apply the distributive property.

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

Apply the distributive property.

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

Subtract $x^{2} \ln (3)$ from both sides of the equation.

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

Add $4 \ln (2)$ to both sides of the equation.

$x^{2} \ln (2) - x^{2} \ln (3) = - 4 \ln (3) + 4 \ln (2)$

Factor $x^{2}$ out of $x^{2} \ln (2) - x^{2} \ln (3)$ .

Factor $x^{2}$ out of $- x^{2} \ln (3)$ .

$x^{2} \ln (2) + x^{2} (- 1 \ln (3)) = - 4 \ln (3) + 4 \ln (2)$

Factor $x^{2}$ out of $x^{2} \ln (2) + x^{2} (- 1 \ln (3))$ .

$x^{2} (\ln (2) - 1 \ln (3)) = - 4 \ln (3) + 4 \ln (2)$

Rewrite $- 1 \ln (3)$ as $- \ln (3)$ .

$x^{2} (\ln (2) - \ln (3)) = - 4 \ln (3) + 4 \ln (2)$

Divide each term in $x^{2} (\ln (2) - \ln (3)) = - 4 \ln (3) + 4 \ln (2)$ by $\ln (2) - \ln (3)$ and simplify.

Divide each term in $x^{2} (\ln (2) - \ln (3)) = - 4 \ln (3) + 4 \ln (2)$ by $\ln (2) - \ln (3)$ .

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

Simplify the left side.

Cancel the common factor of $\ln (2) - \ln (3)$ .

Cancel the common factor.

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

Divide $x^{2}$ by $1$ .

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

Simplify the right side.

Combine the numerators over the common denominator.

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

Factor $4$ out of $- 4 \ln (3) + 4 \ln (2)$ .

Factor $4$ out of $- 4 \ln (3)$ .

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

Factor $4$ out of $4 \ln (2)$ .

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

Factor $4$ out of $4 (- \ln (3)) + 4 \ln (2)$ .

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

Cancel the common factor of $- \ln (3) + \ln (2)$ and $\ln (2) - \ln (3)$ .

Reorder terms.

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

Cancel the common factor.

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

Divide $4$ by $1$ .

$x^{2} = 4$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

Rewrite $4$ as $2^{2}$ .

$x = \pm \sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 2$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 2$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2, - 2$

Solve for x 2^(x^2-4)=3^(x^2-4)

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