tinh
4^2.x-1=7.3^2
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\(3^{x+2}+4\cdot3^{x+1}=7\cdot3^6\)
\(3\cdot3^{x+1}+4\cdot3^{x+1}=7\cdot3^6\)
\(\left(3+4\right)\cdot3^{x+1}=7\cdot3^6\)
\(7\cdot3^{x+1}=7\cdot3^6\)
x + 1 = 6
x = 6 - 1 = 5
Vậy x = 5
\(7\cdot3^{x-1}-3^{x+2}=540\)
=>\(7\cdot3^x\cdot\dfrac{1}{3}-3^x\cdot9=540\)
=>\(3^x\left(\dfrac{7}{3}-9\right)=540\)
=>\(3^x\cdot\dfrac{-20}{3}=540\)
=>\(3^x=-540:\dfrac{20}{3}=-540\cdot\dfrac{3}{20}=-27\cdot3=-81\)
=>\(x=log_3\left(-81\right)\)
42 . x -1 = 7.32
16 . x - 1= 7.9
16 . x -1 = 63
16 . x = 63 + 1
16 . x = 64
x = 64 : 16
x = 4
42.x-1=7.32
16.x-1=7.9
16.x-1=63
16.x=63+1
16.x=64
x=64:16
x=4
\(7\cdot3^{x-1}-3^{x+2}=-540\)
\(\Leftrightarrow3^x\left[7\cdot\left(-3\right)-3^2\right]=-540\)
\(\Leftrightarrow3^x\cdot\left(-30\right)=-540\)
\(\Leftrightarrow3^x=18\)
Ta có : \(7.3^{x-1}-3^{x+2}=-540\)
\(\Leftrightarrow\left(7-3^3\right).3^{x-1}=-540\)
\(\Leftrightarrow\left(7-27\right).3^{x-1}=-540\)
\(\Leftrightarrow-20.3^{x-1}=-540\)
\(\Leftrightarrow3^{x-1}=\left(-540\right):\left(-20\right)\)
\(\Leftrightarrow3^{x-1}=27\)
\(\Leftrightarrow3^{x-1}=3^3\Rightarrow x-1=3\Rightarrow x=4\)
Vậy \(x=4\)
Lời giải:
1.
$3^{x+2}+4.3^{x+1}=7.3^6$
$3^{x+1}.3+4.3^{x+1}=7.3^6$
$3^{x+1}(3+4)=7.3^6$
$3^{x+1}.7=7.3^6$
$\Rightarrow 3^{x+1}=3^6$
$\Rightarrow x+1=6$
$\Rightarrow x=5$
2.
$5^{x+4}-3.5^{x+3}=2.5^{11}$
$5^{x+3}.5-3.5^{x+3}=2.5^{11}$
$5^{x+3}(5-3)=2.5^{11}$
$2.5^{x+3}=2.5^{11}$
$\Rightarrow 5^{x+3}=5^{11}$
$\Rightarrow x+3=11$
$\Rightarrow x=8$
3.
$4^{x+3}-3.4^{x+1}=13.4^{11}$
$4^{x+1}.4^2-3.4^{x+1}=13.4^{11}$
$4^{x+1}.16-3.4^{x+1}=13.4^{11}$
$13.4^{x+1}=13.4^{11}$
$\Rightarrow 4^{x+1}=4^{11}$
$\Rightarrow x+1=11$
$\Rightarrow x=10$
\(4^2x-1=7.3^2\)
<=>\(16x-1=7.9\)
<=>\(16x-1=63\)
<=>\(16x=64\)
<=>\(x=4\)
42 . x - 1 = 7 . 32
42 . x - 1 = 7 . 9
42 . x - 1 = 63
16 . x = 63 + 1
16 . x = 64
x = 64 : 16
x = 4
Hoc tot!!!