DZIAŁANIA NA LOGARYTMACH – rozwiązanie zadań

Zadanie 1

1. (x + 3) − (x − 1) = 2 − 8

(x + 3) − (x − 1) = 16 − 8

1   x + 3 > 0

x > −3

2   x − 1 > 0

x > 1

$$\operatorname{}\frac{x + 3}{x - 1} = \operatorname{}\frac{16}{8}$$

funkcja roznowartosciowa

$$\frac{x + 3}{x - 1} = 2$$

2x − 2 = x + 3

2x − x = 5

x = 5

zestawienie zalozen

D : x ∈ (1, +∞)

1. log(3x+4) +  log(x+8) = 2

log(3x+4)(x+8) = 2

1   3x + 4 > 0

3x > −4

$$x > - \frac{4}{3}$$

2   x + 8 > 0

x > −8

log(3x² + 24x + 4x + 32)=log100

funkcja roznowartosciowa

3x² + 24x + 4x + 32 = 100

3x² + 28x + 32 = 100

zestawienie zalozen

$$x \in ( - \frac{4}{3}, + \infty)$$

1. x − x³ + 2 = 0

x − 3x + 2 = 0

1   x > 0

x = t

t² − 3t + 2 = 0

=9 − 4 * 2 = 1

$$\sqrt{} = 1$$

t₁ = 2

x = 2

funkcja roznowartosciowa

x = 9

x = 9

t₂ = 1

x = 1

funkcja roznowartosciowa

x = 3

x = 3

1. $4 - \log x = 3\sqrt{\log x}$

$$- \log x - 3\sqrt{\log x} + 4 = 0$$

$$\log x + 3\sqrt{\log x} - 4 = 0$$

1   x > 0

$$\mathbf{2}\text{\ \ \ }\sqrt{\log x} > 0$$

3   x ≠ 1

$$\sqrt{\log x} = t$$

t ≥ 0

t² + 3t − 4 = 0

=9 + 16 = 25

$$\sqrt{} = 5$$

t₁ = −4

$$\sqrt{\log x} = - 4$$

−4  nie spelnia zalozenia

t₂ = 1

$$\sqrt{\log x} = 1$$

logx = log10

x = 10

t ≥ 0

logx ≥ 0

logx ≥ log1

x ≥ 1

zestawienie zalozen

D : x ∈ ⟨1 , +∞)

1. x + x + x = 7

1   x > 0

$$\frac{\operatorname{}x}{4} + \frac{\operatorname{}x}{2} + \operatorname{}x = 7\ \ |*4$$

x + 2x + 4x = 28

7x = 28   |:7

x = 4

x = 16

x = 16

1. x^logx + 10x^−logx = 11

1   x > 0

$$x^{\log x} + \frac{10}{x^{\log x}} = 11$$

x^logx = t

$$t + \frac{10}{t} - 11 = 0$$

t² + 10 − 11t = 0

t² − 11t + 10 = 0

=121 − 40 = 81

$$\sqrt{} = 9$$

$$t_{1} = \frac{11 - 9}{2} = 1$$

x^logx = 1

x^logx = x⁰

logx = 0

logx = log1

x = 1

$$t_{2} = \frac{11 + 9}{2} = 10$$

x^logx = 10

logx^logx = log10

logx * logx = 1

x = 1

logx = 1   ∨    logx = −1

$$\log x = \log 10\ \ \ \vee \ \ \ \log x = \log\frac{1}{10}$$

$$x = 10\ \ \ \vee \ \ \ x = \frac{1}{10}$$

1. $2^{\frac{3}{\operatorname{}x}} = \frac{1}{64}$

x > 0

$$2^{\frac{3}{\operatorname{}x}} = 2^{- 6}$$

funkcja roznowartosciowa

$$\frac{3}{\operatorname{}x} = - 6\ \ \ |*\operatorname{}x$$

3 = −6x  |  : ( − 6)

$$- \frac{1}{2} = \operatorname{}x$$

$$\operatorname{}x = - \frac{1}{2}$$

$$\operatorname{}x = \operatorname{}3^{- \frac{1}{2}}$$

$$x = 3^{- \frac{1}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$

1. 3x > 1

3x > 1  |  : 3

$$x > \frac{1}{3}$$

1.

(2x−1) ∈ (0,1)

0 < 2x − 1 < 1

1 < 2x < 2  |  : 2

$$\frac{1}{2} < x < 1$$

$$x \in \left( \frac{1}{2},1 \right)$$

2.

(2x−1) ∈ (1,+∞)

2x − 1 > 1

2x > 2

x > 1

x ∈ (1, +∞)

1. $\left\{ \begin{matrix} \operatorname{}2 + \operatorname{}2 = - \frac{3}{2} \\ \operatorname{}x + \operatorname{}y = - 3 \\ \end{matrix} \right.\ $

1   x > 0

2   y > 0

3   x ≠ 1

4   y ≠ 1

xy = 2⁻³

$$\operatorname{}\text{xy} = \operatorname{}\frac{1}{8}$$

funkcja roznowartosciowa

$$xy = \frac{1}{8}$$

$$y = \frac{1}{8x}$$

$$\operatorname{}2 + \operatorname{}2 = - \frac{3}{2}$$

$$\operatorname{}2 = \frac{\operatorname{}2}{\operatorname{}\frac{1}{8x}} = \frac{\operatorname{}2}{\underset{0}{} - \operatorname{}{8x}} = \frac{\operatorname{}2}{- \operatorname{}{8x}} = \frac{- \operatorname{}2}{\operatorname{}8 + \underset{1}{}} = \frac{- \operatorname{}2}{\operatorname{}8 + 1} = \frac{\operatorname{}2}{\operatorname{}{2^{3} + 1}} = \frac{- \operatorname{}2}{3\operatorname{}2 + 1}$$

$$\operatorname{}2 - \frac{\operatorname{}2}{3\operatorname{}2 + 1} = - \frac{3}{2}$$

2 = t

$$t - \frac{t}{3t + 1} = - \frac{3}{2}\ \ \ |*(3t + 1)$$

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

$$3t^{2} + \frac{9}{2}t + \frac{3}{2} = 0\ \ |*2$$

6t² + 9t + 3 = 0  |  : 3

2t² + 3t + 1 = 0

=1

$$\sqrt{} = \sqrt{1} = 1$$

t₁ = −1

2 = −1

$$\operatorname{}2 = \operatorname{}\frac{1}{x}$$

$$2 = \frac{1}{x}\ \ |*x$$

2x = 1  |   : 2

$$x_{1} = \frac{1}{2}$$

$$t_{2} = - \frac{1}{2}$$

$$\operatorname{}2 = - \frac{1}{2}$$

$$\operatorname{}2 = \operatorname{}{- \frac{1}{2}}$$

$$x^{- \frac{1}{2}} = 2$$

$$\frac{1}{\sqrt{x}} = 2$$

$$\frac{1}{x} = 4\ \ |*x$$

4x = 1

$$x_{2} = \frac{1}{4}$$

$$y = \frac{1}{8x}$$

$$y_{1} = \frac{1}{8x_{1}} = \frac{1}{8*\frac{1}{2}} = \frac{1}{4}$$

$$y_{2} = \frac{1}{8x_{2}} = \frac{1}{8*\frac{1}{4}} = \frac{1}{2}$$

$$\left\{ \begin{matrix} x_{1} = \frac{1}{2} \\ y_{1} = \frac{1}{4} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} x_{2} = \frac{1}{4} \\ y_{2} = \frac{1}{2} \\ \end{matrix} \right.\ $$

Zadanie 2

1. 17²⁴⁰ = 17⁴⁰² = 40² = 800

2. $125^{\operatorname{}16} = 25^{\frac{3}{2}\operatorname{}16} = 25^{\log 25*16\frac{3}{2}} = 16\frac{3}{2} = 4^{3} = 64$

3. $\operatorname{}5*\operatorname{}27 = \frac{1}{\operatorname{}9}*\operatorname{}27 = \frac{\operatorname{}27}{\operatorname{}9} = \frac{\operatorname{}3^{3}}{\operatorname{}3^{2}} = \frac{3\operatorname{}3}{2\operatorname{}3} = \frac{\operatorname{}3}{4\operatorname{}3} = \frac{1}{4}$

Zadanie 3

1. $\operatorname{}a = \sqrt{5}$

$$\operatorname{}\frac{a}{\sqrt{b}} = \operatorname{}\frac{a}{\sqrt{b}} = \frac{\operatorname{}\frac{a}{\sqrt{b}}}{\operatorname{}{\sqrt{a}*\sqrt{b}}} = \frac{\operatorname{}a - \operatorname{}\sqrt{b}}{\operatorname{}\sqrt{a} + \operatorname{}\sqrt{b}} = \frac{\operatorname{}a - \operatorname{}{({b)}^{\frac{1}{2}}}}{\operatorname{}{(a)}^{\frac{1}{2}} + \operatorname{}{(b)}^{\frac{1}{2}}} = \frac{\operatorname{}a - \frac{1}{2}\operatorname{}b}{\frac{1}{2}\operatorname{}a + \frac{1}{2}\operatorname{}b} = \frac{\sqrt{5} - \frac{1}{2}}{\frac{1}{2}\sqrt{5} + \frac{1}{2}}$$

1. x^x = a²x

a ∈ (0,1) ∪ (1, +∞)

x^x = a²x

x * x = a² + x

x = 2a + x

a = 1

x = 2 + x

x − x − 2 = 0

x = t

t² − t − 2 = 0

=1 + 4 * 2 = 9

$$\sqrt{} = 3$$

t₁ = 2

x = 2

x = a²

funkcja roznowartosciowa

x = a²

zalozenie x > 0

t₂ = −1

x = −1

x = x⁻¹

funkcja roznowartosciowa

$$x = \frac{1}{a}$$

2. ${(\sqrt{x})}^{\operatorname{}{x - 1}} = 5$

$$x^{\frac{1}{2}(\operatorname{}{x - 1)}} = 5$$

$$\operatorname{}x^{\frac{1}{2}(\operatorname{}{x - 1)}} = \operatorname{}5$$

$$\left( \frac{1}{2}\operatorname{}x - \frac{1}{2} \right)\operatorname{}x = 1$$

x = t

$$\left( \frac{1}{2}t - \frac{1}{2} \right)t = 1$$

$$\frac{1}{2}t^{2} - \frac{1}{2}t - 1 = 0\ \ \ |*2$$

t² − t − 2 = 0

=1 + 8 = 9

$$\sqrt{} = 3$$

t₁ = −1

x = −1

x = 5⁻¹

$$x = \frac{1}{5}$$

t₂ = 2

x = 2

x = 5²

x = 25

2. 2 * 2 = 2

$$\operatorname{}2 = \frac{\operatorname{}2}{\operatorname{}{2x}} = \frac{\operatorname{}2}{\operatorname{}2 + \operatorname{}x} = \frac{\operatorname{}2}{\operatorname{}2 + 1}$$

$$\operatorname{}2 = \frac{\operatorname{}2}{\operatorname{}{4x}} = \frac{\operatorname{}2}{\operatorname{}4 + \operatorname{}x} = \frac{\operatorname{}2}{2\operatorname{2}{+ 1}}$$

2 = t

$$t*\frac{t}{t + 1} = \frac{t}{2t + 1}$$

$$\frac{t^{2}}{t + 1} = \frac{t}{2t + 1}$$

2t³ + t² = t² + t

2t³ = t

2t³ − t = 0

t(2t²−1) = 0

t = 0

2t² = 1

$$t^{2} = \frac{1}{2}$$

$$t = \frac{\sqrt{2}}{2}\ \ \ \ \vee \ \ \ \ t = - \frac{\sqrt{2}}{2}$$

2 = 0

2 = x⁰

2 = x⁰

2 = 1

rownanie sprzeczne

$$\operatorname{}2 = - \frac{\sqrt{2}}{2}$$

$$\operatorname{}2 = \operatorname{}x^{- \frac{\sqrt{2}}{2}}$$

$$2 = x^{- \frac{\sqrt{2}}{2}}\ \ |* - \frac{2}{\sqrt{2}}$$

$$2^{- \frac{2}{\sqrt{2}}} = x$$

x = 2

$$\operatorname{}2 = \frac{\sqrt{2}}{2}$$

$$2 = x^{\frac{\sqrt{2}}{2}}$$

$$x = 2^{- \sqrt{2}}\ \ \vee \ \ \ x = 2^{\sqrt{2}}$$

1. (2x + 1) > 1

1   2x + 1 > 0

2x > −1

$$x > - \frac{1}{2}$$

1. 

x ∈ (0,1)

funkcja malejaca

(2x + 1)<1

(2x + 1) < x

2x + 1 < x

x < −1

x ∈ ⌀

2.

x ∈ (1,+∞)

(2x+1) > 1

(2x+1) > x

2x + 1 > x

x > −1

x ∈ (1,+∞)

suma rozwiazan

Odpowiedz :     x ∈ (1, +∞)

1. 27 = a    16 = ?

27 = a

3³ = a

$$\operatorname{}16 = \frac{\operatorname{}16}{\operatorname{}6} = \frac{\operatorname{}4^{2}}{\operatorname{}{2*3}} = \frac{\operatorname{}2^{4}}{\operatorname{}{2*3}} = \frac{\operatorname{}2^{4}}{\operatorname{}2 + \operatorname{}3} = \frac{4\operatorname{}2}{\operatorname{}2 + \operatorname{}3}$$

$$\operatorname{}27 = \frac{\operatorname{}27}{\operatorname{}12} = \frac{3\operatorname{}3}{\underset{1}{} + \operatorname{}2} = \frac{3\operatorname{}3}{1 + \operatorname{}2}$$

1. a = 4

$$\log\text{ab} = \frac{1}{\operatorname{}a}$$

$$\frac{1}{\operatorname{}\text{ab}} = 4$$

$$\frac{1}{\underset{1}{} + \operatorname{}b} = \frac{1}{1 + \operatorname{}b} = 4$$

$$\operatorname{}\frac{\sqrt[3]{a}}{\sqrt{b}} = \operatorname{}\sqrt[3]{a} - \operatorname{}\sqrt{b} = \frac{1}{3}\operatorname{}a - \frac{1}{2}\operatorname{}b = \frac{4}{3} - \frac{1}{2}\operatorname{}b = \frac{4}{3} - \frac{1}{2}*\frac{1}{\operatorname{}\text{ab}} = \frac{4}{3} - \frac{1}{2\operatorname{}\text{ab}} = \frac{4}{3} - \frac{1}{2(\operatorname{}a + \underset{1}{}} = \frac{4}{3} - \frac{1}{2(\operatorname{}{a + 1)}} = \frac{4}{3} - \frac{1}{2( - \frac{4}{3} + 1)} = \frac{4}{3} + \frac{3}{2} = \frac{17}{6}$$

$$\frac{1}{4} = 1 + \operatorname{}b$$

$$- \frac{3}{4} = \operatorname{}b$$

$$\operatorname{}{a =} - \frac{4}{3}$$

1. 3^2 − x = 81x

1   x > 0

3² * 3^−x = 81x

$$9*\frac{1}{3^{\operatorname{}x}} = 81x\ \ \ |\ :9$$

$$\frac{1}{3^{\operatorname{}x}} = 9x$$

$$\frac{1}{x} = 9x\ \ \ \ |*x$$

9x² = 1  |  : 9

$$x^{2} = \frac{1}{9}$$

$$x = \frac{1}{3}\ \ \ \ \vee \ \ \ x = - \frac{1}{3} \notin D$$

1. 4^logx = 0, 5 * 10^{1 − log2, 5}

4^logx = 0, 5 * 10 * 10^−log2, 5

$$4^{\log x} = 5*\frac{1}{10^{\log{2,5}}}$$

$$4^{\log x} = \frac{5}{2,5}$$

4^logx = 2

2^2logx = 2¹

funkcja roznowartosciowa

2logx = 1  |  : 2

$$\log x = \frac{1}{2}$$

$$\log x = \log 10^{\frac{1}{2}}$$

$$x = \sqrt{10}$$

Zadanie 4

f(x) = (x³ + 4x² − x − 4)

1 x² − 3 > 0

x² > 3

$$\left( x - \sqrt{3} \right)\left( x + \sqrt{3} \right) > 0$$

$$x = \sqrt{3}\ \ \ \ \vee \ \ \ \ x = - \sqrt{3}$$

$$x \in \left( - \infty, - \sqrt{3} \right) \cup (\sqrt{3}, + \infty)$$

2 x² − 3 ≠ 1

x² ≠ 4

x ≠ 2    ∨    x ≠ −2

3 x³ + 4x² − x − 4 > 0

(x−1)(x²+5x+4) > 0

x = 1

x² + 5x + 4 = 0

=25 − 16 = 9

$$\sqrt{} = 3$$

$$x_{1} = \frac{- 5 - 3}{2} = - 4$$

$$x_{2} = \frac{- 5 + 3}{3} = - 1$$

x ∈ (−4,−1) ∪ (1,+∞)

zestawienie zalozen

$$x \in \left( - 4, - 2 \right) \cup \left( - 2, - \sqrt{3} \right) \cup \left( \sqrt{3},2 \right) \cup (2, + \infty)$$

Zadanie 5

(x − 1) + (x−1) − 2 = 0

(mniejsze od 3)

1   x − 1 > 0

x > 1

2   m ≠ 1

3   m > 1

zal.  (x−1) = t

t² + t − 2 = 0

=1 + 4 * 2 = 9

$$\sqrt{} = 3$$

$$t_{1} = \frac{- 1 + 3}{2} = 1$$

(x−1) = 1

(x−1) = m

x − 1 = m

x = m + 1

m + 1 < 3

m < 2

$$t_{2} = \frac{- 1 - 3}{2} = - 2$$

(x−1) = −2

(x − 1) = m⁻²

$$x - 1 = \frac{1}{m^{2}}$$

$$x = \frac{1}{m^{2}} + 1$$

$$\frac{1}{m^{2}} + 1 < 3$$

$$\frac{1}{m^{2}} < 2$$

1 < 2m²

$$\frac{1}{2} < m^{2}$$

$$m^{2} > \frac{1}{2}$$

2m² > 1

2m² − 1 > 0

$$m^{2} - \frac{1}{2} > 0$$

$$\left( m - \frac{\sqrt{2}}{2} \right)\left( m + \frac{\sqrt{2}}{2} \right) > 0$$

$$m_{0_{1}} = \frac{\sqrt{2}}{2}$$

$$m_{0_{2}} = - \frac{\sqrt{2}}{2}$$

$$m \in \left( - \infty, - \frac{\sqrt{2}}{2} \right) \cup \left( \frac{\sqrt{2}}{2}, + \infty \right)$$

zestawienie z dziedzina

m ∈ (1,+∞)

suma rozwiazan

$$m \in \left( - \infty, - 2 \right) \cup \left( - 2, - \frac{\sqrt{2}}{2} \right) \cup \left( \frac{\sqrt{2}}{2}, + \infty \right)$$