2 mũ x+1- 2 mũ x=32
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a) \(2^{x+1}-2^x=32\)
\(\Rightarrow2^x\left(2-1\right)=2^5\)
\(\Rightarrow2^x.1=2^5\)
\(\Rightarrow x=5\)
b) \(2^{x+1}=2\)
\(\Rightarrow2^{x+1}=2^1\)
\(\Rightarrow x+1=1\)
\(\Rightarrow x=0\)
\(2^{x+1}-2^x=32\)
\(2^x.2-2^x=32\)
\(2^x\left(2-1\right)=32\)
\(2^x=32\)
\(x=5\)
1. 2x=16\(\Rightarrow\)X=4
2. 22x-1=27
\(\Rightarrow\)27=22.4-1
Vậy x =4
\(2^x+2^{x+2}=32.\left(2^2+1\right)\)
\(\Rightarrow2^x+2^{x+2}=32.5\)
\(\Rightarrow2^x+2^{x+2}=160\)
\(\Rightarrow2^x\left(1+4\right)=160\)
\(\Rightarrow2^x.5=160\)
\(\Rightarrow2^x=160:5=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy x = 5
Các bạn giúp mk nha gấp lắm luôn đó nha🤗🤗🤗🤗🤗🤗🤗🤗🤗🤗🤗🤗🤗🤗🤗🤗🤗🤗
TL:
a.\(2^6.2^n=2^{11}\)
\(2^{6+n}=2^{11}\)
\(\Rightarrow n=5\)
b. \(3^7:3^n=3^4\)
\(3^{7-n}=3^4\)
\(\Rightarrow n=3\)
c.\(2^n.32=2^{10}\)
\(2^{n+5}=2^{10}\)
\(\Rightarrow n=5\)
\(\left(X-1\right)^2=16\)
\(\left(X-1\right)^2=4^2\)
X -1 =4
X=5
\(2^{X-1}=32\)
\(2^{X-1}=2^{6-1}\)
X=6
H NHA
a.\(x-1^2=16\) b.\(2^{x-1}=32\)
\(\Leftrightarrow x-1=16\) \(\Leftrightarrow2^{x-1}=2^5\)
\(\Leftrightarrow x=16-1\) \(\Leftrightarrow x-1=5\)
\(\Leftrightarrow x=15\) \(\Leftrightarrow x=5+1\)
Vậy x = 15 \(\Leftrightarrow x=6\)
Vậy x= 6
Cho mk xin 1 tk nhé!
\(\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0^2\)
\(\Leftrightarrow x-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy x = 1/2
\(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left(x-2\right)^2=1^2\)
\(\Leftrightarrow x-2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
Vậy x = 3 hoặc x = 1
\(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-1=-2\)
<=> 2x = -1
<=> x = -0,5
Vậy x = -0,5
\(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
\(\left(x-2\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1+2\\x=-1+2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
Vậy\(x\in\left\{3;1\right\}\)
\(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=-2\)
\(2x=\left(-2\right)+1\)
\(2x=-1\)
\(x=-1\times2\)
\(x=-2\)
\(x\left(\frac{1}{2}\right)^2=\frac{1}{16}\)
\(x\left(\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x\frac{1}{2}=\frac{1}{4}\\x\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}:\frac{1}{2}\\x=-\frac{1}{4}:\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)
\(2^{x+1}-2^x=32\)
\(\Leftrightarrow2^x.2-2^x=2^5\)
\(\Leftrightarrow2^x.\left(2-1\right)=2^5\)
\(\Leftrightarrow2^x.1=2^5\)
\(\Leftrightarrow x=5\)
Vậy : \(x=5\)
\(2^{x+1}-2^x=32\)
\(\Leftrightarrow2^x.2-2^x=2^5\)
\(\Leftrightarrow2^x.\left(2-1\right)=2^5\)
\(\Leftrightarrow2^x=2^5\)
\(\Rightarrow x=5\)
\(2^{x+1}-2^x=32\\ \Rightarrow2^x\cdot2^1-2^x=32\\ \Rightarrow2^x\left(2-1\right)=32\\ \Rightarrow2^x=32:1\\ \Rightarrow2^x=32\\ \Rightarrow2^x=2^5\\ \Rightarrow x=5\)
\(2^{x+1}-2^x=32\\ =>2^x.\left(2^1-1\right)=32\\ =>2^x.1=2^5\\ =>x=5\)