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a, => 2^x = (2^3)^4/(2^4)^3 = 2^12/2^12 = 1 = 2^0
=> x = 0
c, => 4^x = 4^10.(4-3) = 4^10
=> x=10
d, => 2^2.3^x-1 + 2.3^x.9 = 2^2.3^6+2.3^9
=> 2.3^x-1 . (2+3.9) = 2.3^6.(2+3^3)
=> 2.3^x-1 . 27 = 2.3^6 . 27
=> 3^x-1 = 3^6
=> x-1 = 6
=> x = 7
e, => 2^x.(1/3+1/6+2) = 2^11.(2+1/2)
=> 2^x. 5/2 = 2^11. 5/2
=> 2^x = 2^11
=> x = 11
Tk mk nha
a) \(\Leftrightarrow2.\left(\frac{2.3^x}{3}+3^x.3^2\right)=2.3^6\left(2+3^3\right)\)
\(\Leftrightarrow2.\left(\frac{2.3^x+3.3^x.3^2}{3}\right)=2.3^6.29\)
\(\Leftrightarrow2.\left[\frac{3^x.\left(2+3.3^2\right)}{3}\right]=2.3^6.19\)
\(\Leftrightarrow2.3^{x-1}.29=2.3^6.29\Leftrightarrow3^{x-1}.29=\frac{2.3^6.29}{2}=3^6.29\Leftrightarrow3^{x-1}=\frac{3^6.29}{29}=3^6\)
\(\Leftrightarrow3^{x-1}=3^6\Leftrightarrow x-1=6\Leftrightarrow x=6+1=7\)
vậy x=7 . Chọn mình nha
mấy bài sao tương tự nếu ko biết thì nhắn tin mình chỉ típ nha
a)\(\left(-3\right)^{x+3}=-\frac{1}{27}\)
\(\left(-3\right)^{x+3}=\left(-\frac{1}{3}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-\frac{3^0}{3^1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3^{-1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3\right)^{-3}\)
\(\Rightarrow x+3=-3\)
\(\Rightarrow x=-6\)
b)\(\left(-6\right)^{2x+2}=\frac{1}{36}\)
\(\left(-6\right)^{2x+2}=\left(-\frac{1}{6}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-\frac{6^0}{6^1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6^{-1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6\right)^{-2}\)
\(\Rightarrow2x+2=-2\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
c)\(\left(-3\right)^{x+5}=\frac{1}{81}\)
\(\left(-3\right)^{x+5}=\left(-\frac{1}{3}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-\frac{3^0}{3^1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3^{-1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3\right)^{-4}\)
\(\Rightarrow x+5=-4\)
\(\Rightarrow x=-9\)
d)\(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^6\)
\(\left[\left(\frac{1}{3}\right)^2\right]^x=\left[\left(\frac{1}{3}\right)^3\right]^6\)
\(\left(\frac{1}{3}\right)^{2x}=\left(\frac{1}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
e)\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}\right)^3\right]^6\)
\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
1:
\(\Leftrightarrow4\cdot3^x\cdot\dfrac{1}{9}+2\cdot3^x\cdot3=4\cdot3^4+2\cdot3^7\)
\(\Leftrightarrow3^x\cdot\left(\dfrac{4}{9}+6\right)=3^4\cdot\left(4+2\cdot3^3\right)\)
\(\Leftrightarrow3^x=729\)
hay x=6
2: \(\Leftrightarrow3^x\cdot4\cdot\dfrac{1}{3}+3^x\cdot2\cdot9=4\cdot3^6+2\cdot3^9\)
\(\Leftrightarrow3^x\cdot\dfrac{58}{3}=42282\)
=>3x=2187
hay x=7
a, Ta có \(2.3^{x+2}+4.3^{x+1}=3^6.10\)
\(\Rightarrow2.3.3^{x+1}+4.3^{x+1}=3^6.10\)
\(\Rightarrow3^{x+1}.\left(6+4\right)=3^6.10\)
\(\Rightarrow3^{x+1}.10=3^6.10\)
\(\Rightarrow3^{x+1}=3^6\)
\(\Rightarrow x+1=6\)
\(\Rightarrow x=5\)
b,\(\left(\frac{1}{3}+\frac{1}{6}\right).2^{x+4}-2^x=2^{13}-2^{16}\)
\(\Rightarrow\frac{1}{2}.2^{x+4}-2^x=2^{13}.\left(1-2^3\right)\)
\(\Rightarrow2^{x+3}-2^x=2^{13}.\left(1-2^3\right)\)
\(\Rightarrow2^x.\left(2^3-1\right)=2^{13}.\left(1-2^3\right)\)
\(\Rightarrow2^x.\left(2^3-1\right)=-2^{13}.\left(2^3-1\right)\)
\(\Rightarrow2^x=2^{-13}\)
\(\Rightarrow x=-13\)
A ) 2 . 3x+2 + 4 . 33+1 = 36 . 10
2 . 3x . 9 + 4 . 3x . 3 = 729 .10
18 . 3x + 12 . 3x = 243 . 3 . 10
30 . 3x = 243 . 30
3x = 243
x = 5
1: \(5\cdot3^x=5\cdot3^4\)
nên \(3^x=3^4\)
hay x=4
2: \(7\cdot4^x=7\cdot4^3\)
nên \(4^x=4^3\)
hay x=3
3: \(8\cdot7^x=8\cdot7^6\)
nên \(7^x=7^6\)
hay x=6
2.3x+2+4.3x+1=10.36
2.3.3x+1+4.3x+1=10.36
6.3x+1+4.3x+1=10.36
10.3x+1=10.36
=>3x+1=36
=>x+1=6 =>x=5
Ai thấy đúng cho mình nha!
\(2.3^{x+2}+4.3^{x+1}=10.3^6\)
\(2.3^{x+2}+2^2.3^{x+1}=2.5.3^6\)
\(2.3^{x+1}\left(3+2\right)=2.5.3^6\)
\(2.3^{x+1}.5=2.5.3^6\)
\(\Rightarrow x+1=6\Rightarrow x=5\)