Logarithmic equations and inequalities

Simple Logarithmic Equations II

Submitted by michaela.bailova on Sat, 10/12/2024 - 18:56

Solutions to Logarithmic Inequalities

Submitted by michaela.bailova on Wed, 04/17/2024 - 18:09

Simple Logarithmic Equations I

Submitted by michaela.bailova on Fri, 04/05/2024 - 12:38

2000011302

Level:

Find the solution of the equation

\log_{16} x + \log_{4} x + \log_{2} x = 7

x = 16

x = 4

x = \frac{1}{2}

x = 2

2000011301

Level:

Let

x \in (0; 1) \cup (1; + \infty)

. Find the value of

m \in R

2 \log_{m} x = \frac{3}{2} \log_{2} x

m = 2^{\frac{4}{3}}

m = 2^{\frac{3}{4}}

m = \frac{3}{4}

m = \frac{4}{3}

2010010109

Level:

Solve the following inequality.

\log_{0.5} (x + 2) < \log_{0.5} 8

x \in (6; \infty)

x \in [6; \infty)

x \in (- \infty; 6)

x \in (0; 6)

2010010108

Level:

Find the solution set of the following inequality.

\log_{\frac{1}{3}} (x^{2} - 5 x) \geq \log_{\frac{1}{3}} 6

[- 1; 0) \cup (5; 6]

(- 1; 0) \cup (5; 6)

(- 1; 6)

[- 1; 6]

2010010107

Level:

Solve the following equation.

\log_{2} x^{3} \cdot \log_{2} \sqrt[3]{x} + \log_{2} \frac{1}{x} = 6

x_{1} = 8

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

x_{1} = 2

x_{2} = 3

x_{1} = - 8

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

x_{1} = \frac{1}{8}

x_{2} = 4

2010010106

Level:

Which of the following statements about the given equation is true?

\log_{2} (x - 2)^{2} = 4 - \frac{2}{\log_{2} (x - 2)}

The equation has exactly one solution.

The solution set consists of two prime numbers.

The solution set is the empty set.

None of the above statements is true.

2010010105

Level:

Solve.

3^{2 x} = 5

x = \frac{1}{2} \log_{3} 5

x = 2 \log_{3} 5

x = \log_{3} 5^{2}

The equation has no solution.

Logarithmic equations and inequalities

Simple Logarithmic Equations II

Solutions to Logarithmic Inequalities

Simple Logarithmic Equations I

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2000011301

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