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\(\frac{1}{8}.16^n=2^n\)
\(16^n=2^n:\frac{1}{8}\)
\(16^n=2^n.8\)
\(16^n=2^n.2^3\)
\(\left(2^4\right)^n=2^{n+3}\)
\(2^{4n}=2^{n+3}\)
\(\Rightarrow4n=n+3\)
\(4n-n=3\)
\(3n=3\)
\(n=1\)
\(KL:n=1\)
CHÚC BN HỌC TỐT!!!!!
\(\frac{1}{8}.16^x=2^x\\ \Leftrightarrow\frac{2^x}{16^x}=\frac{1}{8}\\ \Leftrightarrow\left(\frac{2}{16}\right)^x=\frac{1}{8}\\ \Leftrightarrow\left(\frac{1}{8}\right)^x=\frac{1}{8}\Leftrightarrow x=1\)
Vậy x = 1
\(\frac{1}{8}.16^x=2^x\)
\(\Rightarrow\frac{1}{8}=\frac{2^x}{16^x}\)
\(\Rightarrow\frac{1}{8}=\left(\frac{2}{16}\right)^x\)
\(\Rightarrow\frac{1}{8}=\left(\frac{1}{8}\right)^x\)
\(\Rightarrow\left(\frac{1}{8}\right)^x=\left(\frac{1}{8}\right)^1\)
\(\Rightarrow x=1\)
Vậy \(x=1.\)
Chúc bạn học tốt!
c, \(\frac{-32}{-2^n}=4\)
\(\Rightarrow-2^n=-32:4\)
\(\Rightarrow-2^n=-8\)
\(\Rightarrow-2^n=-2^3\Rightarrow n=3\)
d, \(\frac{8}{2^n}=2\)
\(\Rightarrow2^n=8:2\)
\(\Rightarrow2^n=4\)
\(\Rightarrow2^n=2^2\Rightarrow n=2\)
e, \(\frac{25^3}{5^n}=25\)
\(\Rightarrow5^n=25^3:25\)
\(\Rightarrow5^n=25^2\)
\(\Rightarrow5^n=5^4\Rightarrow n=4\)
i , \(8^{10}:2^n=4^5\)
\(\Rightarrow2^n=8^{10}:4^5\)
\(\Rightarrow2^n=\left(2^3\right)^{10}:\left(2^2\right)^5\)
\(\Rightarrow2^n=2^{30}:2^{10}\)
\(\Rightarrow2^n=2^{20}\Rightarrow n=20\)
k, \(2^n.81^4=27^{10}\)
\(\Rightarrow2^n=27^{10}:81^4\)
\(\Rightarrow2^n=\left(3^3\right)^{10}:\left(3^4\right)^4\)
\(\Rightarrow2^n=3^{30}:3^{16}\)
\(\Rightarrow2^n=3^{14}\)
\(\Rightarrow2^n=4782969\)Không chia hết cho 2 nên ko có Gt n thỏa mãn
Câu 1 :
\(\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^5}\)
= \(\frac{5^{32}.2^{86}.5^{43}}{\left(-2\right)^{87}.5^{15}}\)
= \(\frac{5^{72}.\left(-2\right)^{86}}{\left(-2\right)^{87}.5^{75}}\)
= \(\frac{1}{-2}\)
Câu 2 :
\(\frac{5^4.18^4}{125.9^5.16}\)
= \(\frac{5^4.2^4.3^8}{5^3.3^{10}.2^4}\)
= \(\frac{5}{3^2}\)
= \(\frac{5}{9}\)
Câu 3 :
\(\frac{9^{18}.2^{29}}{8^9.27^{12}}\)
= \(\frac{3^{36}.2^{29}}{2^{27}.3^{36}}\)
= \(2^2\)
= 4
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) \(32^{-n}.16^n=2048\)
\(\Leftrightarrow\left(2^5\right)^{-n}.\left(2^4\right)^n=2^{11}\)
\(\Leftrightarrow2^{-5n}.2^{4n}=2^{11}\)
\(\Leftrightarrow2^{-n}=2^{11}\)
\(\Leftrightarrow n=-11\)
\(\frac{1}{8}.16^n=2^n\)
\(\frac{16^n}{8}=2^n\)
\(\Rightarrow\frac{2^{4n}}{2^3}=2^n\)
\(2^{4n-3}=2^n\)
\(4n-3=1\)
\(\Rightarrow n=1\)
Vậy n = 1
\(\frac{1}{8}.16^n=2^n\Rightarrow\frac{1}{8}=\frac{2^n}{16^n}\Rightarrow\frac{1}{8}=\left(\frac{2}{16}\right)^n\Rightarrow\frac{1}{8}=\left(\frac{1}{8}\right)^n\Rightarrow n=1\)
vậy n=1