Question

Решите уравнение:

$1) 27^{3x-1} = frac{1}{9} ^{x} 2) 5^{3x} +3* 5^{3x-2} =140 3) 4^{x} - 2^{x} -12=0 4) 11^{x} = 2^{x}$

5)(3/5)^(x2-x)=(5/3)^-2

Answer 1

$1)\ \ 27^{3x-1}=frac{1}{9}^{x}\ \ (3^3)^{3x-1}=(3^{-2})^{x}\ \ 3^{9x-3}=3^{-2x}\ \ Rightarrow 9x-3=-2x\ \ 9x-3+2x=0\ \ 11x=3\ \ boxed{x=frac{3}{11}}$

$2)\ \ 5^{3x}+3 cdot 5^{3x-2}=140\ \ 5^{3x-2} cdot 5^{2}+3 cdot 5^{3x-2}=140\ \ 5^{3x-2} (5^{2}+3)=140\ \ 5^{3x-2} cdot 28=140\ \ 5^{3x-2}=140:28\ \ 5^{3x-2}=5^{1}\ \ Rightarrow 3x-2=1\ \ 3x=1+2\ \ 3x=3\ \ x=3:3\ \ boxed{x=1}$

$3)\ \ 4^{x}-2^{x}-12=0\ \ (2^{2})^{x}-2^{x}=12\ \ 2^{2x}-2^{x}=12\ \ 2^{x}(2^{2}-1)=12\ \ 2^{x} cdot 3=12\ \ 2^{x}=12:3\ \ 2^{x}=4\ \ 2^{x}=2^{2}\ \ Rightarrow boxed{x=2}$

$5)\ \ (frac{3}{5})^{x^{2}-x}=(frac{5}{3})^{-2}\ \ (frac{3}{5})^{x^{2}-x}=[(frac{3}{5})^{-1}]^{-2}\ \ (frac{3}{5})^{x^{2}-x}=(frac{3}{5})^{2}\ \ Rightarrow x^{2}-x=2\ \ x^{2}-x-2=0\ \ Delta=b^{2}-4ac\ \ a=1;~b=-1;~c=-2\ \ Delta=(-1)^{2}-4cdot1cdot(-2)\ \ Delta=1+8=9\ \ x_{1,2}=frac{-b pm sqrt{Delta}}{2a}\ \x_{1,2}=frac{-(-1) pm sqrt{9}}{2cdot1}\ \ x_{1,2}=frac{1 pm 3}{2}\ \ x_{1}=frac{1+3}{2}=frac{4}{2}=2\ \ x_{2}=frac{1-3}{2}=frac{-2}{2}=-1\ \ x in { -1;~2 }$