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\(\left(\frac{1}{3}\right)^{2x-1}-\frac{1}{3^2}=-\frac{2}{27}\)
=> \(\left(\frac{1}{3}\right)^{2x-1}=-\frac{2}{27}+\frac{1}{9}\)
=> \(\left(\frac{1}{3}\right)^{2x-1}=\frac{1}{27}\)
=> \(\left(\frac{1}{3}\right)^{2x-1}=\left(\frac{1}{3}\right)^3\)
=> 2x - 1 = 3
=> 2x = 3 + 1
=> 2x = 4
=> x = 4/2 = 2
64 . 4^x = 16^8
=> 4^3 . 4^x = (4^2) ^8
=> 4^3 . 4^x = 4^16
=> 4^x = 4^16 : 4^ 3
=> 4^x = 4^ 13
=> x = 13
9^x = 27^4
=> ( 3^2) ^ x = ( 3^3 ) 4
=>3^2.x = 3^ 12
=> 2.x = 12
=> x = 12 : 2
=> x = 6
Câu 1 = 4^13
Câu 2 = 9^6
Câu cuối mik không làm được ;)
k mik nhé
BÀI 1 dễ òi nên k giải nữa nha, chỉ cần ghép các số ( 1;2;3 ) số đầu, liên tiếp dần là đc nha bạn.
Bài 2:
\(8^4\cdot16^5=\left(2^3\right)^4\cdot\left(2^4\right)^5=2^{12}\cdot2^{20}=2^{32}\)
\(5^{40}\cdot125^7\cdot625^3=5^{40}\cdot\left(5^3\right)^7\cdot\left(5^4\right)^3=5^{40}\cdot5^{21}\cdot5^{12}=5^{73}\)
\(27^4\cdot81^{10}=\left(3^3\right)^4\cdot\left(3^4\right)^{10}=3^{12}\cdot3^{40}=3^{52}\)
\(10^3\cdot100^5\cdot1000^4=10^3\cdot\left(10^2\right)^5\cdot\left(10^3\right)^4=10^3\cdot10^{10}\cdot10^{12}=10^{25}\)
a, \(2^x-15=17\)
\(\Rightarrow2^x=17+15\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
b, \(\left(7x-11\right)^3=2^5.5^2+200\)
\(\Rightarrow\left(7x-11\right)^3=32.25+200\)
\(\Rightarrow\left(7x-11\right)^3=1000\)
\(\Rightarrow\left(7x-11\right)^3=10^3\)
\(\Rightarrow7x-11=10\)
\(\Rightarrow7x=10+11\)
\(\Rightarrow7x=21\)
\(\Rightarrow x=21:7\)
\(\Rightarrow x=3\)
c, \(x^{10}=1^x\)
\(\Rightarrow x\in\left\{1;0\right\}\)
\(2^x-15=17\)
\(\Rightarrow2^x=17+15\)
\(\Rightarrow2^x=32=2^4\)
\(\Rightarrow x=4\)
\(\left(7x-11\right)^3=2^5.5^2+200\)
Phần này mk ko bt làm đâu
\(x^{10}=1^x\)
\(\Rightarrow\)\(x^{10}=1\)
\(\Rightarrow x=1\)
a) \(2^{4x+1}-8^{x+2}=0\)\(\Leftrightarrow2^{4x+1}-2^{3\left(x+2\right)}=0\)
\(\Leftrightarrow2^{4x+1}-2^{3x+6}=0\)\(\Leftrightarrow2^{4x+1}=2^{3x+6}\)
\(\Leftrightarrow4x+1=3x+6\)\(\Leftrightarrow4x-3x=6-1\)\(\Leftrightarrow x=5\)
Vậy \(x=5\)
b) \(3^2.9^{2x}=27^{x+3}\)\(\Leftrightarrow3^2.3^{2.2x}=3^{3\left(x+3\right)}\)\(\Leftrightarrow3^2.3^{4x}=3^{3x+9}\)
\(\Leftrightarrow3^{2+4x}=3^{3x+9}\)\(\Leftrightarrow2+4x=3x+9\)\(\Leftrightarrow4x-3x=9-2\)\(\Leftrightarrow x=7\)
Vậy \(x=7\)
c) \(8^{2x}.64^2=16^{x+4}\)\(\Leftrightarrow2^{3.2x}.2^{6.2}=2^{4\left(x+4\right)}\)\(\Leftrightarrow2^{6x}.2^{12}=2^{4\left(x+4\right)}\)
\(\Leftrightarrow2^{6x+12}=2^{4x+16}\)\(\Leftrightarrow6x+12=4x+16\)\(\Leftrightarrow6x-4x=16-12\)
\(\Leftrightarrow2x=4\)\(\Leftrightarrow x=2\)
Vậy \(x=2\)
a) \(3^x=81\)
\(3^x=3^4\)
\(\Rightarrow x=4\)
b) \(2^x.16=128\)
\(2^x=128:16\)
\(2^x=8\)
\(2^x=2^3\)
\(\Rightarrow x=3\)
c) \(3^x:9=27\)
\(3^x=27.9\)
\(3^x=243\)
\(3^x=3^5\)
\(\Rightarrow x=5\)
d) \(x^4=x\)
\(\Rightarrow x=0\)hoac \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
e) \(\left(2x+1\right)^3=27\)
\(\left(2x+1\right)^3=3^3\)
\(\Rightarrow2x+1=3\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
f) \(\left(x-2\right)^2=\left(x-2\right)^4\)
\(\left(x-2\right)^2-\left(x-2\right)^4=0\)
\(\left(x-2\right)^2-\left(x-2\right)^2.\left(x-2\right)^2=0\)
\(\left(x-2\right)^2\left[1-\left(x-2\right)^2\right]=0\)
\(\left(x-2\right)^2\left(1-x+2\right)\left(1+x-2\right)=0\)
\(\Rightarrow\left(x-2\right)^2=0\)hoac \(\orbr{\begin{cases}3-x=0\\x-1=0\end{cases}}\)
\(\Rightarrow x-2=0\)hoac \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
\(\Rightarrow x=2\)hoac \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
a) \(3^x=81\Leftrightarrow3^x=3^4\Rightarrow x=4\)
b)\(2^x\times16=128\Leftrightarrow2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
c) \(3^x\div9=27\Leftrightarrow3^x\div3^2=3^3\Rightarrow x=5\)
d) \(x^4=x\Leftrightarrow x=1\)
e) \(\left(2x+1\right)^3=27\Leftrightarrow\left(2x+1\right)^3=3^3\Rightarrow2x+1=3 \)
\(\Rightarrow2x=3+1\Leftrightarrow2x=4\Rightarrow x=2\)
F)
=> 2^x = 16 x^3 = 27
=> 2^x = 2^4 =>x^3 = 3^3
=> x = 4 => x = 3
Vậy x = 4 Vậy x = 3
X =4
x =9
nhớ k cho mk