Упр.40.13 ГДЗ Мордкович 10-11 класс (Алгебра)

1. а) $$3^x-3^{x+3}=-78$$;
   б) $$5^{2x-1}-5^{2x-3}=4{,}8$$;
   в) $$2\cdot\left(\frac{1}{7}\right)^{3x+7}-7\cdot\left(\frac{1}{7}\right)^{3x+8}=49$$;
   г) $$\left(\frac{1}{3}\right)^{5x-1}+\left(\frac{1}{3}\right)^{5x}=\frac{4}{9}$$.

а) $$3^x-3^{x+3}=-78$$

Вынесем $$3^x$$ за скобки:

$$3^x(1-3^3)=-78$$

$$3^x(1-27)=-78$$

$$-26\cdot 3^x=-78$$

$$3^x=3$$

$$x=1$$

б) $$5^{2x-1}-5^{2x-3}=4{,}8$$

Вынесем $$5^{2x-3}$$ за скобки:

$$5^{2x-3}(5^2-1)=4{,}8$$

$$5^{2x-3}\cdot 24=4{,}8$$

$$5^{2x-3}=\frac{4{,}8}{24}=0{,}2=\frac{1}{5}$$

$$5^{2x-3}=5^{-1}$$

$$2x-3=-1$$

$$2x=2$$

$$x=1$$

в) $$2\cdot \left(\frac{1}{7}\right)^{3x+7}-7\cdot \left(\frac{1}{7}\right)^{3x+8}=49$$

Вынесем $$\left(\frac{1}{7}\right)^{3x+7}$$ за скобки:

$$\left(\frac{1}{7}\right)^{3x+7}\left(2-7\cdot \frac{1}{7}\right)=49$$

$$\left(\frac{1}{7}\right)^{3x+7}(2-1)=49$$

$$\left(\frac{1}{7}\right)^{3x+7}=49$$

$$\left(\frac{1}{7}\right)^{3x+7}=\left(\frac{1}{7}\right)^{-2}$$

$$3x+7=-2$$

$$3x=-9$$

$$x=-3$$

г) $$\left(\frac{1}{3}\right)^{5x-1}+\left(\frac{1}{3}\right)^{5x}=\frac{4}{9}$$

Вынесем $$\left(\frac{1}{3}\right)^{5x}$$ за скобки:

$$\left(\frac{1}{3}\right)^{5x}\left(\left(\frac{1}{3}\right)^{-1}+1\right)=\frac{4}{9}$$

$$\left(\frac{1}{3}\right)^{5x}(3+1)=\frac{4}{9}$$

$$4\left(\frac{1}{3}\right)^{5x}=\frac{4}{9}$$

$$\left(\frac{1}{3}\right)^{5x}=\frac{1}{9}=\left(\frac{1}{3}\right)^2$$

$$5x=2$$

$$x=\frac{2}{5}=0{,}4$$

Ответ

а) $$1$$; б) $$1$$; в) $$-3$$; г) $$0{,}4$$.