Tìm n biết:2b+3=3k
B=3+3 mũ 2+3 mũ 3+...+3 mũ 51
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(Nếu chỗ \(3k=3^n\) thì tham khảo nhé)
\(B=3+3^2+3^3+...+3^{51}\\ 3B=3^2+3^3+3^4+...+3^{52}\\ 3B-B=3^2+3^3+3^4+...+3^{52}-3-3^2-3^3-...-3^{51}\\ 2B=3^{51}-3\\ B=\dfrac{3^{51}-3}{2}\\ 2B+3=\dfrac{3^{51}-3}{2}.2+3=3^{51}=3^n\Rightarrow n=51\)
\(A=3+3^2+3^3+...+3^{10}\)
\(3A=3\cdot\left(3+3^2+3^3+...+3^{10}\right)\)
\(3A=3^2+3^3+3^4+...+3^{11}\)
\(3A-A=3^2+3^3+3^4+...+3^{11}-3-3^2-3^3-...-3^{10}\)
\(2A=3^{11}-3\)
Nên ta có:
\(2A+3=3^n\)
\(\Rightarrow3^{11}-3+3=3^n\)
\(\Rightarrow3^n=3^{11}\)
\(\Rightarrow n=11\)
Bài 6 :
a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)
c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)
d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)
Bài 7 :
a) \(3^x+3^{x+2}=9^{17}+27^{12}\)
\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)
\(\Rightarrow10.3^x=3^{34}+3^{36}\)
\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)
\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)
b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)
\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)
\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)
c) Bài C bạn xem lại đề
d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)
\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)
\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)
\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)
\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)
\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)
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
b: Ta có: \(2^{x+3}+2^x=144\)
\(\Leftrightarrow2^x\cdot9=144\)
\(\Leftrightarrow2^x=16\)
hay x=4
\(B=3+3^2+3^3+...+3^{51}\)
\(\Rightarrow B=3\left(1+3^1+3^2+...+3^{50}\right)\)
\(\Rightarrow B=3.\dfrac{3^{50+1}-1}{3-1}\)
\(\Rightarrow B=\dfrac{3\left(3^{51}-1\right)}{2}\)
Ta có :
\(2B+3=3n\)
\(\Rightarrow2.\dfrac{3\left(3^{51}-1\right)}{2}+3=3n\)
\(\Rightarrow3n-3=3\left(3^{51}-1\right)\)
\(\Rightarrow3\left(n-1\right)=3\left(3^{51}-1\right)\)
\(\Rightarrow n-1=3^{51}-1\)
\(\Rightarrow n=3^{51}-1+1\)
\(\Rightarrow n=3^{51}\)
\(B=3+3^2+3^3+...+3^{51}\)
\(3B=3^2+3^3+...+3^{52}\)
\(3B-B=3^2+3^3+3^4+...+3^{52}-3-3^2-...-3^{51}\)
\(2B=3^{52}-3\)
\(B=\dfrac{3^{52}-3}{2}\)
Mà:
\(2B+3=3^n\)
\(\Rightarrow2\cdot\dfrac{3^{52}-3}{2}+3=3^n\)
\(\Rightarrow3^{52}-3+3=3^n\)
\(\Rightarrow3^{52}=3^n\)
\(\Rightarrow n=52\)