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第 10 章复习练习
https://query.libretexts.org/%E7%AE%80%E4%BD%93%E4%B8%AD%E6%96%87/%E4%B8%AD%E7%BA%A7%E4%BB%A3%E6%95%B0_(OpenStax)/10%3A_%E6%8C%87%E6%95%B0%E5%92%8C%E5%AF%B9%E6%95%B0%E5%87%BD%E6%95%B0/10.0E%3A_%E7%AC%AC_10_%E7%AB%A0%E5%A4%8D%E4%B9%A0%E7%BB%83%E4%B9%A0
2。 $\begin{array}{l}{\log _{5} 8+2 \log _{5} a+6 \log _{5} b} {+\log _{5} c-3 \log _{5} d}\end{array}$ 4。 \(\begin{array}{l}{\frac{1}{3}\left(\log _{6} 7+2 \log _{6} x-1-3 \log _{6} y\right.} {-5 \l...2。 $\begin{array}{l}{\log _{5} 8+2 \log _{5} a+6 \log _{5} b} {+\log _{5} c-3 \log _{5} d}\end{array}$ 4。 $\begin{array}{l}{\frac{1}{3}\left(\log _{6} 7+2 \log _{6} x-1-3 \log _{6} y\right.} {-5 \log _{6} z )}\end{array}$ $\log (x-1)-\log (3 x+5)=-\log x$ $\log _{4}(x-2)+\log _{4}(x+5)=\log _{4} 8$ 3。 $\begin{array}{l}{\frac{1}{4}\left(\log _{2} 5+3 \log _{2} x-4-2 \log _{2} y\right.} {-7 \log _{2} z )}\end{array}$ 求解 $x$ ： $\log _{7}(x+2)+\log _{7}(x-3)=\log _{7} 24$
Capítulo 10 Exercícios de revisão
https://query.libretexts.org/Idioma_Portugues/Algebra_intermediaria_(OpenStax)/10%3A_Fun%C3%A7%C3%B5es_exponenciais_e_logar%C3%ADtmicas/10.0E%3A_Cap%C3%ADtulo_10_Exerc%C3%ADcios_de_revis%C3%A3o
Nos exercícios a seguir, para cada conjunto de pares ordenados, determine se ela representa uma função e, em caso afirmativo, é a função um a um. Nos exercícios a seguir, determine se cada gráfico é o...Nos exercícios a seguir, para cada conjunto de pares ordenados, determine se ela representa uma função e, em caso afirmativo, é a função um a um. Nos exercícios a seguir, determine se cada gráfico é o gráfico de uma função e, em caso afirmativo, é um para um. 2. $\begin{array}{l}{\log _{5} 8+2 \log _{5} a+6 \log _{5} b} {+\log _{5} c-3 \log _{5} d}\end{array}$ $\log (x-1)-\log (3 x+5)=-\log x$ Resolva para $x$ : $\log _{7}(x+2)+\log _{7}(x-3)=\log _{7} 24$
Sura ya 10 Mazoezi Mapitio
https://query.libretexts.org/Kiswahili/Algebra_ya_kati_(OpenStax)/10%3A_Kazi_za_kielelezo_na_za_Logarithmic/10.0E%3A_Sura_ya_10_Mazoezi_Mapitio
Katika mazoezi yafuatayo, kwa kila seti ya jozi zilizoamriwa, onyesha ikiwa inawakilisha kazi na ikiwa ni hivyo, ni kazi moja kwa moja. 2. \(\begin{array}{l}{\log _{5} 8+2 \log _{5} a+6 \log _{5} b} {...Katika mazoezi yafuatayo, kwa kila seti ya jozi zilizoamriwa, onyesha ikiwa inawakilisha kazi na ikiwa ni hivyo, ni kazi moja kwa moja. 2. $\begin{array}{l}{\log _{5} 8+2 \log _{5} a+6 \log _{5} b} {+\log _{5} c-3 \log _{5} d}\end{array}$ $\log (x-1)-\log (3 x+5)=-\log x$ 3. $\begin{array}{l}{\frac{1}{4}\left(\log _{2} 5+3 \log _{2} x-4-2 \log _{2} y\right.} {-7 \log _{2} z )}\end{array}$ Tatua kwa $x$ : $\log _{7}(x+2)+\log _{7}(x-3)=\log _{7} 24$
Chapitre 10 Exercices de révision
https://query.libretexts.org/Francais/Alg%C3%A8bre_interm%C3%A9diaire_(OpenStax)/10%3A_Fonctions_exponentielles_et_logarithmiques/10.0E%3A_Chapitre_10_Exercices_de_r%C3%A9vision
Dans les exercices suivants, utilisez la propriété de produit des logarithmes pour écrire chaque logarithme sous la forme d'une somme de logarithmes. 2. \(\begin{array}{l}{\log _{5} 8+2 \log _{5} a+6 ...Dans les exercices suivants, utilisez la propriété de produit des logarithmes pour écrire chaque logarithme sous la forme d'une somme de logarithmes. 2. $\begin{array}{l}{\log _{5} 8+2 \log _{5} a+6 \log _{5} b} {+\log _{5} c-3 \log _{5} d}\end{array}$ $\log (x-1)-\log (3 x+5)=-\log x$ 3. $\begin{array}{l}{\frac{1}{4}\left(\log _{2} 5+3 \log _{2} x-4-2 \log _{2} y\right.} {-7 \log _{2} z )}\end{array}$ Résolvez pour $x$ : $\log _{7}(x+2)+\log _{7}(x-3)=\log _{7} 24$
الفصل 10 تمارين المراجعة
https://query.libretexts.org/%D8%A7%D9%84%D9%84%D8%BA%D8%A9_%D8%A7%D9%84%D8%B9%D8%B1%D8%A8%D9%8A%D8%A9/%D8%A7%D9%84%D8%AC%D8%A8%D8%B1_%D8%A7%D9%84%D9%85%D8%AA%D9%88%D8%B3%D8%B7_(OpenStax)/10%3A_%D8%A7%D9%84%D8%AF%D9%88%D8%A7%D9%84_%D8%A7%D9%84%D8%A3%D8%B3%D9%8A%D8%A9_%D9%88%D8%A7%D9%84%D9%84%D9%88%D8%BA%D8%A7%D8%B1%D9%8A%D8%AA%D9%85%D9%8A%D8%A9/10.0E%3A_%D8%A7%D9%84%D9%81%D8%B5%D9%84_10_%D8%AA%D9%85%D8%A7%D8%B1%D9%8A%D9%86_%D8%A7%D9%84%D9%85%D8%B1%D8%A7%D8%AC%D8%B9%D8%A9
2. $\begin{array}{l}{\log _{5} 8+2 \log _{5} a+6 \log _{5} b} {+\log _{5} c-3 \log _{5} d}\end{array}$ 4. \(\begin{array}{l}{\frac{1}{3}\left(\log _{6} 7+2 \log _{6} x-1-3 \log _{6} y\right.} {-5 \l...2. $\begin{array}{l}{\log _{5} 8+2 \log _{5} a+6 \log _{5} b} {+\log _{5} c-3 \log _{5} d}\end{array}$ 4. $\begin{array}{l}{\frac{1}{3}\left(\log _{6} 7+2 \log _{6} x-1-3 \log _{6} y\right.} {-5 \log _{6} z )}\end{array}$ $\log (x-1)-\log (3 x+5)=-\log x$ $\log _{4}(x-2)+\log _{4}(x+5)=\log _{4} 8$ 3. $\begin{array}{l}{\frac{1}{4}\left(\log _{2} 5+3 \log _{2} x-4-2 \log _{2} y\right.} {-7 \log _{2} z )}\end{array}$ حل لـ $x$ : $\log _{7}(x+2)+\log _{7}(x-3)=\log _{7} 24$