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ta có công thức như sau :
\(a^{-x}=?\)
lời giải công thức này như sau :
\(a^{-x}=\left(\frac{1}{a}\right)^x\)
vậy bài cũng gải tương tự
\(32^{-x}.16^x=\left(\frac{1}{32}\right)^x.\left(16^x\right)\)
\(=\left(\frac{16}{32}\right)^x=\left(\frac{1}{2}\right)^x=2^{-x}\)
mà \(2048=2^{11}\)
\(\Rightarrow-x=11\)
\(\Leftrightarrow x=-11\)
vậy \(x=-11\)
\(\Rightarrow\)\(\left(\frac{1}{32}\right)^x\cdot16^x=2048\)
\(\Rightarrow\)\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^{-11}\)
\(\Rightarrow\)\(x=-11\)
a) \(\dfrac{1}{9}.27^n=3^n\)
\(\Leftrightarrow\dfrac{1}{9}=3^n:27^n\)
\(\Leftrightarrow\dfrac{1}{9}=\left(\dfrac{3}{27}\right)^n\)
\(\Leftrightarrow\dfrac{1}{9}=\left(\dfrac{1}{9}\right)^n\)
\(\Leftrightarrow n=1\)
b) \(3^{-2}.3^4.3^n=3^7\)
\(\Leftrightarrow3^2.3^n=3^7\)
\(\Leftrightarrow3^n=3^7:3^2\)
\(\Leftrightarrow3^n=3^5\)
\(\Leftrightarrow n=5\)
c) \(32^{-n}.16^n=2048\)
\(\Leftrightarrow\left(2^5\right)^{-n}.\left(2^4\right)^n=2^{11}\)
\(\Leftrightarrow2^{-5n}.2^{4n}=2^{11}\)
\(\Leftrightarrow2^{-n}=2^{11}\)
\(\Leftrightarrow n=-11\)
a)\(32^{-n}\cdot16^n=2048\)
\(\left(2^5\right)^{-n}\cdot\left(2^4\right)^n\)=2048
\(2^{-5n}\cdot2^{4n}\)=\(2^{11}\)
\(2^{-5n+4n}=2^{11}\)
\(2^{-x}=2^{11}\)
\(\Rightarrow x=-11\)
b)\(2^{-1}\cdot2^n+4\cdot2^n=9\cdot2^5\)
\(\frac{1}{2}\cdot2^n+4\cdot2^n=288\)
\(2^n\left(\frac{1}{2}+4\right)=288\)
\(2^n\cdot\frac{9}{2}=288\)
\(2^n=288:\frac{9}{2}\)
\(2^n=64\)
\(2^n=2^6\)
\(\Rightarrow n=6\)
a) 32-n . 16n = 2048
\(\frac{1}{32n}\) . 16n = 2048
\(\frac{1}{2^n.16^n}\) . 16n = 2048
\(\frac{1}{2^n}\) = 2048
2-n = 2048
2-n = 211
\(\Rightarrow\) -n = 11
\(\Rightarrow\) n = -11
Vậy n = -11
\(7.4^{x+1}-5.4^{x+1}=2048\)
\(\Rightarrow4^{x+1}.\left(7-5\right)=2048\)
\(\Rightarrow4^{x+1}.2=2048\)
\(\Rightarrow4^{x+1}=2048:2\)
\(\Rightarrow4^{x+1}=1024\)
\(\Rightarrow4^{x+1}=4^5\)
\(\Rightarrow x+1=5\)
\(\Rightarrow x=5-1\)
\(\Rightarrow x=4\)
Vậy \(x=4.\)
Chúc bạn học tốt!
x = -11 do ban