3^4.3^x=3^7
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a) \(3\cdot3^4\cdot3^5=3^{10}\)
b) \(7^3:7^2:7=7^{3-2-1}=7^0\)
c) \(\left(x^4\right)^3=x^{12}\)
a, \(\left(\frac{1}{2}-\frac{1}{3}\right)\cdot6^x+6^{x+2}=6^{10}+6^7\)
\(\Leftrightarrow\frac{1}{6}\cdot6^x+6^x\cdot6^2=6^{10}+6^7\)
\(\Leftrightarrow6^{x-1}\left(1+6^3\right)=6^7\left(6^3+1\right)\)
\(\Leftrightarrow6^{x-1}=6^7\Leftrightarrow x-1=7\)
\(\Leftrightarrow x=8\)
b, \(\left(\frac{1}{2}-\frac{1}{6}\right)\cdot3^{x+4}-4\cdot3^x=3^{16}-4\cdot3^{13}\)
\(\Leftrightarrow\frac{1}{3}\cdot3^{x+4}-4\cdot3^x=3^{13}\left(3^3-4\right)\)
\(\Leftrightarrow3^x\cdot3^3-4\cdot3^x=3^{13}\left(3^3-4\right)\)
\(\Leftrightarrow3^x\left(3^3-4\right)=3^{13}\left(3^3-4\right)\)
\(\Leftrightarrow3^x=3^{13}\Leftrightarrow x=13\)
a. x=8
b. x=13
còn cách tính thì mình quên rồi vì minh học cái này lâu lắm rồi ko nhớ đc.
Lời giải:
$3^{x-1}+4.3^{x-2}=\frac{7}{243}$
$\Leftrightarrow 3. 3^{x-2}+4.3^{x-2}=\frac{7}{243}$
$\Leftrightarrow 3^{x-2}(3+4)=\frac{7}{243}$
$\Rightarrow 3^{x-2}=\frac{1}{243}=3^{-5}$
$\Rightarrow x-2=-5$
$\Rightarrow x=-3$
\(3^{x-1}+4.3^{x-2}=\frac{7}{243}\)
\(\Rightarrow3^1.3^{x-2}+4.3^{x-2}=\frac{7}{243}\)
\(\Rightarrow3^{x-2}.\left(3^1+4\right)=\frac{7}{243}\)
\(\Rightarrow3^{x-2}.7=\frac{7}{243}\)
\(\Rightarrow3^{x-2}=\frac{7}{243}:7\)
\(\Rightarrow3^{x-2}=\frac{1}{243}\)
\(\Rightarrow3^{x-2}=3^{-5}\)
\(\Rightarrow x-2=-5\)
\(\Rightarrow x=\left(-5\right)+2\)
\(\Rightarrow x=-3\)
Vậy \(x=-3.\)
Chúc bạn học tốt!
a) \(\left(x+5\right).6-7=29\)
\(\Rightarrow\left(x+5\right).6=29+7\)
\(\Rightarrow\left(x+5\right).6=36\)
\(\Rightarrow\left(x+5\right)=36:6=6\)
\(\Rightarrow x=6-5=1\)
b) \(5x-3x=12-3.2\)
\(\Rightarrow2x=6\Rightarrow x=6:2=3\)
c) \(\dfrac{1}{4}.3< x< \dfrac{51}{50}.4\)
\(\Rightarrow\dfrac{3}{4}< x< \dfrac{102}{25}\)
Bài 1:
a) Ta có: \(\dfrac{7^4\cdot3-7^3}{7^4\cdot6-7^3\cdot2}\)
\(=\dfrac{7^3\cdot\left(7\cdot3-1\right)}{7^3\cdot2\left(7\cdot3-1\right)}\)
\(=\dfrac{1}{2}\)
c) Ta có: \(E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
\(\Leftrightarrow\dfrac{1}{3}\cdot E=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E-\dfrac{1}{3}\cdot E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)
\(\Leftrightarrow E\cdot\dfrac{2}{3}=1-\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E=\dfrac{3-\dfrac{3}{3^{101}}}{2}=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)
a ) 5^n = 125
5^n = 5^3
n = 3
b ) 3^4 . 3^n = 3^7
3^(4 + n ) =3^7
4 + n = 7
n = 7 - 4
n = 3
\(3^4.3^x=3^7\\ \Rightarrow4+x=7\\ \Rightarrow x=3.\)
\(3^4.3^x=3^7\)
\(81.3^x=2187\)
\(3^x=2187:81\)
\(3^x=27\)
\(\Rightarrow3^2=27\)
\(\Rightarrow x=2\)