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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 \(x=-11\)
\(\Rightarrow\)\(\left(\frac{1}{32}\right)^x\cdot16^x=2048\)
\(\Rightarrow\)\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^{-11}\)
\(\Rightarrow\)\(x=-11\)
\(\frac{2030-x}{15}+\frac{2041-x}{13}+\frac{2048-x}{11}+\frac{1961-x}{9}=0\)
\(\Leftrightarrow\frac{2030-x}{15}-1+\frac{2041-x}{13}-2+\frac{2048-x}{11}-3+\frac{1961-x}{9}+6=0\)
\(\Leftrightarrow\frac{2015-x}{15}+\frac{2015-x}{13}+\frac{2015-x}{11}+\frac{2015-x}{9}=0\)
\(\Leftrightarrow\left(2015-x\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)
Mà \(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\ne0\)
\(\Rightarrow2015-x=0\Leftrightarrow x=2015\)
\(A=1+3+3^2+3^3+...+3^{101}\)
\(3A=3+3^2+3^3+3^4+...+3^{101}\)
\(3A-A=\left(3+3^2+3^3+3^4+...+3^{101}\right)-\left(1+3+3^2+3^3+...+3^{100}\right)\)
\(2A=3^{101}-1\)
\(A=\left(3^{101}-1\right):2\)
Thu gọn tổng sau:
A=1+3+32+33+...+3100
B= 2100-299-298-297-...-22-2
C= 3100-399+398-397-...+32-3+1
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
a, \(\frac{1}{9}.27^n=3^n\Leftrightarrow\frac{1}{9}.3^{3.n}=3^n\Leftrightarrow\frac{1}{3^2}=3^n:3^{3n}\Leftrightarrow\frac{1}{3^2}=3^{n-3n}=3^{2n}\)
=> 3^2n . 3^2 = 1 => 3^( 2n + 2) = 3^0 => 2n + 2 = 0 => 2n = - 2 => n = - 1
b, 3^-2.3^4 .3^n = 3^ 7 => 3^ ( -2 + 4 + n) = 3^7 => 3^ (n+ 2) = 3^7 => n + 2 = 7 => n = 5
Thấy: `2 xx 2048 = 4096 = 2^12 = 2^6 xx 2^6 = 64 xx 64`.
`-> x = 64 (64^64 = 2^2048)`.
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