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a) \(4^n=4096\Rightarrow4^n=4^6\Rightarrow n=6\)
b) \(5^n=15625\Rightarrow5^n=5^6\Rightarrow n=6\)
c) \(6^{n+3}=216\Rightarrow6^{n+3}=6^3\Rightarrow n+3=3\Rightarrow n=0\)
d) \(x^2=x^3\Rightarrow x^3-x^2=0\Rightarrow x^2\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
e) \(3^{x-1}=27\Rightarrow3^{x-1}=3^3\Rightarrow x-1=3\Rightarrow x=4\)
f) \(3^{x+1}=9\Rightarrow3^{x+1}=3^2\Rightarrow x+1=2\Rightarrow x=1\)
g) \(6^{x+1}=36\Rightarrow6^{x+1}=6^2\Rightarrow x+1=2\Rightarrow x=1\)
h) \(3^{2x+1}=27\Rightarrow3^{2x+1}=3^3\Rightarrow2x+1=3\Rightarrow2x=2\Rightarrow x=1\)
i) \(x^{50}=x\Rightarrow x^{50}-x=0\Rightarrow x\left(x^{49}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{49}-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x^{49}=1=1^{49}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
4n = 4096
4n = 212
n = 12
5n = 15625
5n = 56
n = 6
6n+3 = 216
6n+3 = 23.33
6n+3 = 63
n + 3 = 3
a) \(3^2.x+2^3.x=51\)
\(\Leftrightarrow x\left(3^2+2^3\right)=51\)
\(\Leftrightarrow17x=51\)
\(\Leftrightarrow x=3\)
Vậy
b) \(6^2.2-\left(84-3^2.x\right):7=69\)
\(\Leftrightarrow\left(84-3^2.x\right):7=3\)
\(\Leftrightarrow84-3^2.x=21\)
\(\Leftrightarrow3^2.x=63\)
\(\Leftrightarrow x=7\)
Vậy
Bài 1:
2\(x\) = 4
2\(^x\) = 22
\(x=2\)
Vậy \(x=2\)
Bài 2:
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
Ta có 2x= 4y-1 hay 2x=22(y-1) => x=2(y-1)=2y-2 (1)
27y=3x+8 hay 33y = 3x+8 => 3y=x+8 (2)
thay (1) vào (2) ta được 3y= 2y-2+8
=> 3y-2y=6 => y=6
ta có x=2y-2=2. 6 -2=10
x=10 ; y=6
a)\(3^x=9\Leftrightarrow x^x=3^2\Leftrightarrow x=2\)
b)\(6^{x-1}=36\Leftrightarrow6^{x-1}=6^2\Leftrightarrow x-1=2\Leftrightarrow x=3\)
c)\(5^x=125\Leftrightarrow5^x=5^3\Leftrightarrow x=3\)
d)\(3^{2x+1}=27\Leftrightarrow3^{2x+1}=3^3\Leftrightarrow2x+1=3\Leftrightarrow2x=2\Leftrightarrow x=1\)
e) \(3^{x+1}=9\Leftrightarrow3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=1\)
f) \(x^{50}=x\Leftrightarrow\hept{\begin{cases}x=1\\x=0\end{cases}}\)
\(325-5\left[4^3-\left(27-5^2\right)\cdot1^{18}\right]\\ =325-5\left[64-\left(27-25\right)\cdot1\right]\\ =325-5\left(64-2\right)=325-5\cdot62=325-310=15\)
1) \(\left(-27\right).\left(-28+128\right)=-27.100=-2700\)
2a)\(\left(x-3\right)\left(2x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
b) \(\left(2x-1\right)^2=81\)
\(\sqrt{\left(2x-1\right)^2}=9\)
\(\left|2x-1\right|=9\)
\(\left[{}\begin{matrix}2x-1=9\\2x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)
c) \(\left(2m+5\right)^3=-27\)
\(\sqrt[3]{\left(2m+5\right)^3}=-3\)
\(2m+5=-3\)
\(m=-4\)
d) \(\left(3x-2\right)^3=64\)
tương tự câu c
\(\left(x+1\right)^3=27\)
\(\left(x+1\right)^3=3^3\)
\(\Rightarrow x+1=3\)
\(x=2\)
\(\left(x+1\right)^3=27\)
\(< =>\left(x+1\right)^3=3.3.3=3^3\)
\(< =>x+1=3< =>x=3-1=2\)
\(\left(2x+3\right)^3=9.81\)
\(< =>\left(2x+3\right)^3=9.9.9\)
\(< =>\left(2x+3\right)^3=9^3\)
\(< =>2x+3=9< =>2x=6\)
\(< =>x=\frac{6}{2}=3\)