Задание 2.7

Question

2.7. Решите уравнение:
\begin{enumerate}[label=\arabic*)]
\item $\frac{1}{9} \cdot \sqrt{3^{3 x-1}}=81^{-\frac{3}{4}}$; 
\item $4^{x} \cdot 3^{x+1}=0,25 \cdot 12^{3 x-1}$;
\item $4 \cdot 2^{\cos x}=\sqrt{8}$;
\item $0,25 \cdot 2^{x^{2}}=\sqrt[3]{0,25 \cdot 4^{2 x}}$;
\item $5^{x-1}=10^{x} \cdot 2^{-x} \cdot 5^{x+1}$;
\item $\sqrt[3]{9^{2 x+1}}=\frac{3}{\sqrt[5]{3}}$.
\end{enumerate}

Gimli · Accepted Answer

\begin{enumerate}[label=\arabic*)]
\item 
$$\frac{1}{9} \cdot {\sqrt{3^{3 x-1}}}=81^{-\frac{3}{4}} $$
$$3^{-2}\cdot 3^\frac{3x-1}{2}=81^{-\frac{3}{4}}$$
$$3^{-\frac{4}{2}}\cdot 3^{\frac{3 x-1}{2}}=\left(3^{4}\right)^{-\frac{3}{4}}$$ 
$$3^{\frac{-4+ 3 x-1}{2}}=\left(3^{4}\right)^{-\frac{3}{4}}$$ 
$$\frac{3 x-5}{2}=-3 $$
$$3 x-5=-6$$
$$3 x=-1$$
$$x=-\frac{1}{3}.$$

\item 
$$4^{x} \cdot 3^{x+1}=4^{-1}\cdot 4^{3 x-1} \cdot 3^{3 x-1}$$
$$4^{x} \cdot 3^{x+1}=4^{3 x-2} \cdot 3^{3 x-1}$$
$$4^{x} \cdot 3^{x}\cdot 3=\frac{4^{3 x}}{4^{2}} \cdot \frac{3^{3 x}}{3}$$
$$12^{x}\cdot 3=\frac{12^{3 x}}{48}$$
$$3\cdot48\cdot 12^{x}=12^{3 x}$$
$$3\cdot48\cdot 12^{x}-12^{3 x}=0$$
$$12^{x}\left(144-12^{2 x}\right) =0$$
$$12^{x} =0\qquad x \in \emptyset$$
$$144-12^{2 x}=0$$
$$12^{2 x} =12^{2 }$$
$$2x=2$$
$$x=1$$

\item 
$$4 \times 2^{\cos x}=\sqrt{8}$$
$$2^{2} \cdot 2^{\cos x}=2^{\frac{3}{2}}$$
$$2^{\cos x+2}=2^{\frac{3}{2}}$$
$$\cos x+2=1,5$$
$$\cos x=-\frac{1}{2}$$
$$x=\pm \frac{2 \pi}{3}+2 \pi k, k \in Z$$

\item 
$$0,25 \cdot 2^{x^{2}}=\sqrt[3]{0,25 \cdot 4^{2 x}}$$
$$4^{-1} \cdot 2^{x^{2}}=\left(4^{-1} \cdot\left(4\right)^{2 x}\right)^{\frac{1}{3}}$$
$$2^{-2} \cdot 2^{x^{2}}=\left(2^{-2} \cdot\left(2^{2}\right)^{2 x}\right)^{\frac{1}{3}}$$
$$2^{-2+x^{2}}=2^{(-2+4 x)^{3}}$$

$$2+x^{2}=\frac{4 x-2}{3}$$
$$-6+3 x^{2}=4 x-2$$
$$3 x^{2}-4 x-4=0,  $$
$$D=16+3 \cdot 4 \cdot 4=64$$
$$x_{1}=\frac{4+8}{6}=2$$
$$x_{2}=\frac{4-8}{6}=-\frac{2}{3}$$

\item 
$$5^{x-1}=10^{x} \cdot 2^{-x} \cdot 5^{x+1}$$ 
$$5^{x-1}=5^{x} \cdot 2^{x} \cdot 2^{-x} \cdot 5^{x+1}$$
$$5^{x-1}=5^{x+x+1} \cdot 2^{0}$$
$$5^{x-1}=5^{2 x+1}$$
$$x-1=2 x+1$$
$$x=-2$$

\item 
$$\sqrt[3]{9^{2 x+1}}=\frac{3}{\sqrt[5]{3}}$$
$$\left(\left(3^{2}\right)^{2 x+1}\right)^{\frac{1}{3}}=3^{1}\div 3^{\frac{1}{5}}$$
$$3^{\frac{4 x+2}{3}}=3^{1-\frac{1}{5}}$$
$$\frac{4 x+2}{3}=\frac{4}{5}$$
$$5(4 x+2)=3 \cdot 4$$
$$20 x+10=12$$
$$20 x=2$$
$$x=\frac{1}{10} .$$
\end{enumerate}

Задание 2.7

Answers

Comments

Задание 2.7

Answers

Comments

Add answer