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

а) 3^x — 3^(x + 3) = -78;
б) 5^(2x — 1) — 5^(2x — 3) = 4,8;
в) 2 * (1/7)^(3x + 7) — 7 * (1/7)^(3x + 8) = 49;
г) (1/3)^(5x — 1) + (1/3)^5x = 4/9.

1. $$3^x-3^{x+3}=-78$$

   Вынесем $$3^x$$ за скобки:

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

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

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

   $$3^x=3$$

   $$3^x=3^1 \Rightarrow x=1$$

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

   Вынесем $$5^{2x-3}$$ за скобки:

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

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

   $$5^{2x-3}=\frac{4{,}8}{24}=0{,}2=\frac15$$

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

   $$2x-3=-1$$

   $$2x=2$$

   $$x=1$$

3. $$2\cdot \left(\frac17\right)^{3x+7}-7\cdot \left(\frac17\right)^{3x+8}=49$$

   Вынесем $$\left(\frac17\right)^{3x+7}$$ за скобки:

   $$\left(\frac17\right)^{3x+7}\left(2-7\cdot \frac17\right)=49$$

   $$\left(\frac17\right)^{3x+7}(2-1)=49$$

   $$\left(\frac17\right)^{3x+7}=49$$

   $$\left(\frac17\right)^{3x+7}=\left(\frac17\right)^{-2}$$

   $$3x+7=-2$$

   $$3x=-9$$

   $$x=-3$$

4. $$\left(\frac13\right)^{5x-1}+\left(\frac13\right)^{5x}=\frac49$$

   Вынесем $$\left(\frac13\right)^{5x}$$ за скобки:

   $$\left(\frac13\right)^{5x}\left(\left(\frac13\right)^{-1}+1\right)=\frac49$$

   $$\left(\frac13\right)^{5x}(3+1)=\frac49$$

   $$4\left(\frac13\right)^{5x}=\frac49$$

   $$\left(\frac13\right)^{5x}=\frac19=\left(\frac13\right)^2$$

   $$5x=2$$

   $$x=\frac25=0{,}4$$

Ответ

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