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1)\(8.2^n=128\Rightarrow2^n=128:8\Rightarrow2^n=16\Rightarrow2^n=2^4\Rightarrow n=4\)
2)\(121.11^n=1331\Rightarrow11^n=1331:121\Rightarrow11^n=11\Rightarrow n=1\)
3)\(7^n:49=343\Rightarrow7^n:7^2=7^3\Rightarrow7^n=7^3.7^2\Rightarrow7^n=7^5\Rightarrow n=5\)
nhớ **** cho mình nhé
\(\frac{2}{3}.x=\frac{1}{3}\)1) x-\(\frac{10}{3}\)=\(\frac{7}{15}.\frac{3}{5}\)
x-10/3=7/25
x=7/25+10/3
x=\(\frac{271}{75}\)
2)\(\frac{8}{23}.\frac{46}{24}.x=\frac{1}{3}\)
2/3.x=1/3
x=1/3:2/3
x=1/2
\(\dfrac{2}{49}\) x 121 : \(\dfrac{22}{7}\)
= \(\dfrac{2}{49}\) x 121 x \(\dfrac{7}{22}\)
= \(\dfrac{2}{7\times7}\) \(\times\)\(\dfrac{11\times11\times7}{11\times2}\)
= \(\dfrac{11}{7}\)
b: \(\Leftrightarrow\left[{}\begin{matrix}x+7=11\\x+7=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-18\end{matrix}\right.\)
Ta có \(a=\frac{1}{2}\log_711;b=\log_27\)
Mặt khác : \(\log_{\sqrt[3]{7}}\frac{121}{8}=3\log_7\frac{11^2}{2^3}=3\left(2\log_711-3\log_72\right)=6\log_711-\frac{9}{\log_27}=12a-\frac{9}{b}\)
Vậy \(\log_{\sqrt[3]{7}}\frac{121}{8}=12a-\frac{9}{b}\)
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