[tex]\displaystyle \sf \log_a(b) + \log_b(a) = \sqrt{(2025)} \\\\ \left[log_a(b)\right]^2 + \left[\log_b(a)\right]^2 = \ ? \\\\\\\ \text{Fa\c ca} : \\\\\left[\log_a(b) + \log_b(a)\right]^2=\left(\sqrt{2025}\right)^2 \\\\ \left[log_a(b)\right]^2+\left[\log_b(a)\right]^2+2\cdot \log_a(b)\cdot \log_b(a) = 2025 \\\\\ \text{Pela mudan\c ca de base sabe-se que}: \\\\ \log_b(a) = \frac{\log_a(a)}{\log_a(b)}\\\\ \log_b(a) = \frac{1}{\log_a(b)} \\\\\ \log_a(b)\cdot \log_b(a) = 1[/tex]

[tex]\displaystyle \sf Assim, \\\\ \left[log_a(b)\right]^2+\left[\log_b(a)\right]^2+2\cdot 1 = 2025\\\\ \left[log_a(b)\right]^2+\left[\log_b(a)\right]^2 =2025-2\\\\ \boxed{\sf \ \left[log_a(b)\right]^2+\left[\log_b(a)\right]^2 = 2023 \ }\checkmark[/tex]

letra c)

Verified answer

Resposta:

[log_a (b)]^2 + [log_b (a)]^2 = 2023, alternativa c)

Explicação passo-a-passo:

Solução:

log_a (b) = m

log_b (a) = n

m + n = (√(2025))

(m + n)^2 = ((√(2025)))^2

m^2 + 2mn + n^2 = 2025

m^2 + n^2 = 2025 − 2mn

m^2 + n^2 = 2025 − 2[log_a (b)][log_a (b)]

m^2 + n^2 = 2025 − 2[log_a (b)][(1/(log_b (a)))]

m^2 + n^2 = 2025 − 2(1)

m^2 + n^2 = 2025 − 2

m^2 + n^2 = 2023