IME / ITA ⇒ (Escola Naval - 2004) Logaritmos Tópico resolvido

[ALDRINMOD]

Se [tex3]a[/tex3], [tex3]b[/tex3], [tex3]m[/tex3] e [tex3]n[/tex3] são números reais tais que [tex3]a^2+b^2=341ab[/tex3], [tex3]a \neq 0[/tex3], [tex3]b \neq 0[/tex3], [tex3]\log_3 2=m[/tex3] e [tex3]\log_3 7=n[/tex3], então, o valor da expressão [tex3]\log_3 \frac{[a+b]^2}{64ab}-\log_3 \[\frac{7}{3}\]^2-2[\log_9 2]^2+\log_{\frac{1}{3}} 14[/tex3] é

(A) [tex3]m^2+6n-1[/tex3].
(B) [tex3]{-}\frac{m^2}{2}-7m+2[/tex3].
(C) [tex3]3\frac{n^2}{2}+3m-6n-2[/tex3].
(D) [tex3]\frac{n^2}{2}+6n-1[/tex3].
(E) [tex3]{-}n^2+6m-1[/tex3].

Resposta

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(A)

[triplebig]

[tex3]\begin{array}{rl}
\mbox{S}=&\log_3\frac{(a+b)^2}{64ab}-\log_3\left(\frac{7}{3}\right)^2-2\left(\log_92\right)^2+\log_{\frac{1}{3}}14\\


\text{ }&\text{ }\\
=&\log_3\frac{a^2+b^2+2ab}{64ab}-2\left(\log_37-\log_33\right)-2\cdot\left(\frac{1}{2}\log_32\right)^2+\log_3\left(\frac{1}{14}\right)\\


\text{ }&\text{ }\\
=&\log_3\left(\frac{7}{4}\right)^3-2\cdot(n-1)-\frac{m^2}{2}+\log_31-\left(\log_32+\log_37\right)\\


\text{ }&\text{ }\\
=&3\cdot\left(\log_37-2\log_32\right)-2n+2-\frac{m^2}{2}-n-m\\


\text{ }&\text{ }\\
=&3n-6m-2n+2-\frac{m^2}{2}-n-m\\


\text{ }&\text{ }\\
=&\boxed{-\frac{m^2}{2}-7m+2}
\end{array}[/tex3]

É, a minha deu a letra (B) .