13) \( \log _{5}\left\{\left[\log _{2}\left(3-\log _{4} x^{2}\right)^{2}+1\right]+2\right\}=1 S \)

To solve the equation \( \log _{5}\left\{\left[\log _{2}\left(3-\log _{4} x^{2}\right)^{2}+1\right]+2\right\}=1 \), we start by simplifying it step-by-step. 1. First, the equation \( \log_{5}(A) = 1 \), where \( A = \left[\log_{2}\left(3-\log_{4} x^{2}\right)^{2}+1\right]+2 \), implies: \[ A = 5^1 = 5 \] So we can rewrite it as: \[ \left[\log_{2}\left(3-\log_{4} x^{2}\right)^{2}+1\right]+2 = 5 \] 2. Next, we isolate the logarithm: \[ \log_{2}\left(3-\log_{4} x^{2}\right)^{2}+1 = 5 - 2 = 3 \] Therefore: \[ \log_{2}\left(3-\log_{4} x^{2}\right)^{2} = 3 - 1 = 2 \] 3. We can now exponentiate both sides using base 2: \[ \left(3-\log_{4} x^{2}\right)^{2} = 2^2 = 4 \] 4. Taking the square root of both sides (considering both the positive and negative roots): \[ 3 - \log_{4} x^{2} = 2 \quad \text{or} \quad 3 - \log_{4} x^{2} = -2 \] 5. **Case 1**: \( 3 - \log_{4} x^{2} = 2 \): - Solving gives: \[ \log_{4} x^{2} = 1 \] - Converting the log: \[ x^{2} = 4^1 = 4 \] - Hence, \( x = \pm 2 \). 6. **Case 2**: \( 3 - \log_{4} x^{2} = -2 \): - Solving gives: \[ \log_{4} x^{2} = 5 \] - Converting the log: \[ x^{2} = 4^5 = 1024 \] - Hence, \( x = \pm 32 \). Finally, we have four possible solutions. However, we need to check the logarithmic conditions to ensure they are valid. The values we found are: \[ x = 2, \; x = -2, \; x = 32, \; x = -32 \] These values meet the conditions where \( x > 0 \) and \( x < 0 \) aren’t producing invalid logarithmic arguments. Thus, the valid solutions are: \[ \boxed{2}, \; \boxed{32} \quad (\text{valid for } x > 0) \] And the negative roots are technically valid but need context based on application. Final answers: \[ x = \pm 2, \quad x = \pm 32 \]