Question

Ajutati ma va rog cu C 16 și dacă va rog sa mi scrieți ce formule folosiți în rezolvare sau pas cu pas sa înțeleg. Mulțumesc frumos

Answer 1

 

$\displaystyle\\C16)\\a=\log_9\sqrt{3}+\log_4\sqrt[3]{2}=?\\\\\text{Formule folosite:}\\\\\sqrt{a}=a^{(\frac{1}{2})}\\\\\sqrt[n]{a}=a^{(\frac{1}{n})}\\\\log_aa=1~~~a>0,~a\neq1 \\\\\log_ab^c=c\cdot\log_ab~~~a>0,~a\neq1,~b>0,~c\in R \\\\\log_{a^c}b=\frac{1}{c}\cdot\log_ab~~~a>0,~a\neq1,~b>0,~c\in R-\{0\}$

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$\displaystyle\\\text{Rezolvare:}\\a=\log_9\sqrt{3}+\log_4\sqrt[3]{2}\\\\a=\log_93^\frac{1}{2}+\log_42^\frac{1}{3}\\\\a=\frac{1}{2}\cdot\log_93+\frac{1}{3}\cdot\log_42\\\\a=\frac{1}{2}\cdot\log_{3^2}3+\frac{1}{3}\cdot\log_{2^2}2\\\\a=\frac{1}{2}\cdot\frac{1}{2}\cdot\log_{3}3+\frac{1}{2}\cdot\frac{1}{3}\cdot\log_{2}2\\\\a=\frac{1}{2}\cdot\frac{1}{2}\cdot1+\frac{1}{2}\cdot\frac{1}{3}\cdot1\\\\a=\frac{1}{4}+\frac{1}{6}\\\\a=\frac{3}{12}+\frac{2}{12}\\\\a=\frac{3+2}{12}\\\\\boxed{a=\frac{5}{12}\in Q}$