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a, \(2\cdot2^2\cdot2^3\cdot2^4\cdot...\cdot2^x=1024\)
\(\Leftrightarrow2^{1+2+3+4+...+x}=2^{10}\Leftrightarrow1+2+3+4+...+x=10\)
\(\Rightarrow\left(x+1\right)x\div2=10\Rightarrow\left(x+1\right)x=20\)
Vì : ( x + 1 ) x là hai số tự nhiên liên tiếp \(\Rightarrow x=4\in Z\)
Vậy x = 4
b, \(9.27< 3^x< 243\Leftrightarrow3^5< 3^x< 3^5\)
\(\Rightarrow5< x< 5\Rightarrow x\in\varnothing\)
Vậy \(x\in\varnothing\)
a) Sai đề
b)\(\left(x-2\right)^6=4^3\)
\(\left(x-2\right)^6=4^3=64=2^6=\left(-2\right)^6\)
\(\Rightarrow x-2=2\)hoặc \(x-2=-2\)
\(Th1:x-2=2\Leftrightarrow x=4\)
\(Th2:x-2=-2\Leftrightarrow x=0\)
c)\(\left(x+1\right)^2=3^4\)
\(\left(x+1\right)^2=3^4=81=9^2=\left(-9\right)^2\)
\(\Rightarrow x+1=9\)hoặc \(x+1=-9\)
\(Th1:x+1=9\Leftrightarrow x=8\)
\(Th2:x+1=-9\Leftrightarrow x=-10\)
d) \(3^{x+1}=243\)
\(3^{x+1}=3^5\)
\(\Leftrightarrow x+1=5\Leftrightarrow x=4\)
e)\(2^x=4^{10}=\left(2^2\right)^{10}=2^{20}\Leftrightarrow x=20\)
\(\frac{1}{3}\left(2x-\frac{1}{2}\right)^4=\frac{1}{243}\)
=> \(\left(2x-\frac{1}{2}\right)^4=\frac{1}{243}:\frac{1}{3}=\frac{1}{243}\cdot3=\frac{1}{81}\)
=> \(\left(2x-\frac{1}{2}\right)^4=\left(\pm\frac{1}{3}\right)^4\)
=> \(\hept{\begin{cases}2x-\frac{1}{2}=\frac{1}{3}\\2x-\frac{1}{2}=-\frac{1}{3}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{5}{12}\\x=\frac{1}{12}\end{cases}}\)
(2x-1/2)^4=1/243:1/3
(2x-1/2)=1/243×3/1
(2x-1/2)^4=1/81
(2x-1/2)^4=1/3^4
(2x-1/2)=1/3
2x=1/3+1/2
2x=5/6
x=5/6:2/1
x=5/6×1/2
x=5/12
Bài 1:
\(\text{a) }x.x^2.x^3.x^4.x^5.....x^{49}.x^{50}\)
\(=x^{1+2+3+4+5+...+49+50}\)
\(=x^{\frac{51.50}{2}}\)
\(=x^{1275}\)
\(\text{b) Ta có:}\)
\(4^{15}=\left(2^2\right)^{15}=2^{2.15}=2^{30}\)
\(8^{11}=\left(2^3\right)^{11}=2^{3.11}=2^{33}\)
\(\text{Vì }2^{30}< 2^{33}\text{ nên }4^{15}< 8^{11}\)
Bài 2: Tìm x
\(\left(x-1\right)^4:3^2=3^6\)
\(\Rightarrow\left(x-1\right)^4=3^6\times3^2\)
\(\Rightarrow\left(x-1\right)^4=3^8\)
\(\Rightarrow\left(x-1\right)^4=3^{2.4}\)
\(\Rightarrow\left(x-1\right)^4=\left(3^2\right)^4\)
\(\Rightarrow x-1=9\)
\(\Rightarrow x=10\)
Bài 3 và bài 4 mk làm sau
Bài 1 : a) \(x.x^2.x^3.x^4.....x^{49}.x^{50}=x^{1+2+3+...+49+50}\) (Dễ rồi tự tính)
b) \(\hept{\begin{cases}4^{15}=\left(2^2\right)^{15}=2^{30}\\8^{11}=\left(2^3\right)^{11}=2^{33}\end{cases}}\)Rồi tự so sánh đi
Bài 2 :
\(\left(x-1\right)^4\div3^2=3^6\Leftrightarrow\left(x-1\right)^4=3^8=\left(3^2\right)^4=9^4\Leftrightarrow x-1=9\Leftrightarrow x=10\)
Bài 3 :
\(\hept{\begin{cases}27^{15}=\left(3^3\right)^{15}=3^{45}\\81^{11}=\left(3^4\right)^{11}=3^{44}\end{cases}}\) nt
3x( 1 + 3 + 32) = 36.13
3x. 13 = 36.13
=> 3x= 36
=> x= 6
mk ko hiểu cho lắm