Math Problem Statement

реши

Решить уравнение и неравенство: 1) (1/5)^(2-3x) = 25, 2) 4^x + 2^x - 20 = 0, неравенство (3/4)^x > 1/3, систему уравнений {x-y=4, 5x+y=25}, неравенство (√5)^(x-6) < 1/5 и (2/13)^(x^2-1) >= 1, уравнение 7^(x+1) + 3 * 7^x = 2x^5 + 3 * 2x.

Solution

Let's go through the problems one by one:

1. Solve the equation:

  1. (15)23x=25\left(\frac{1}{5}\right)^{2 - 3x} = 25

We can express 2525 as 525^2, so the equation becomes: (15)23x=(5)2=(15)2\left(\frac{1}{5}\right)^{2 - 3x} = \left(5\right)^2 = \left(\frac{1}{5}\right)^{-2} Equating the exponents: 23x=22 - 3x = -2 Solving for xx: 3x=223x=4x=43-3x = -2 - 2 \quad \Rightarrow \quad -3x = -4 \quad \Rightarrow \quad x = \frac{4}{3}

  1. 4x+2x20=04^x + 2^x - 20 = 0

Let’s set y=2xy = 2^x. Then 4x=(22)x=y24^x = (2^2)^x = y^2, so the equation becomes: y2+y20=0y^2 + y - 20 = 0 This is a quadratic equation. Solving it using the quadratic formula: y=1±124(1)(20)2(1)=1±1+802=1±812=1±92y = \frac{-1 \pm \sqrt{1^2 - 4(1)(-20)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 80}}{2} = \frac{-1 \pm \sqrt{81}}{2} = \frac{-1 \pm 9}{2} Thus: y=1+92=4ory=192=5y = \frac{-1 + 9}{2} = 4 \quad \text{or} \quad y = \frac{-1 - 9}{2} = -5 Since y=2xy = 2^x must be positive, y=4y = 4, so 2x=42^x = 4 gives x=2x = 2.

2. Solve the inequality:

(34)x>13\left(\frac{3}{4}\right)^x > \frac{1}{3}

Taking the logarithm of both sides (since 34<1\frac{3}{4} < 1, the inequality will reverse): xln(34)<ln(13)x \ln\left(\frac{3}{4}\right) < \ln\left(\frac{1}{3}\right) Since ln(34)\ln\left(\frac{3}{4}\right) is negative, we divide by it and reverse the inequality: x>ln(13)ln(34)x > \frac{\ln\left(\frac{1}{3}\right)}{\ln\left(\frac{3}{4}\right)} Calculating the result gives the value for xx.

3. Solve the system of equations:

x - y = 4 \\ 5x + y = 25 \end{cases}$$ Adding the two equations to eliminate $$y$$: $$(x - y) + (5x + y) = 4 + 25 \quad \Rightarrow \quad 6x = 29 \quad \Rightarrow \quad x = \frac{29}{6}$$ Substituting $$x = \frac{29}{6}$$ into the first equation: $$\frac{29}{6} - y = 4 \quad \Rightarrow \quad y = \frac{29}{6} - 4 = \frac{29}{6} - \frac{24}{6} = \frac{5}{6}$$ Thus, $$x = \frac{29}{6}$$ and $$y = \frac{5}{6}$$. ### 4. Solve the inequality: 1) $$\left(\sqrt{5}\right)^{x - 6} < \frac{1}{5}$$ Express $$\frac{1}{5}$$ as $$5^{-1}$$, and $$\sqrt{5}$$ as $$5^{1/2}$$: $$\left(5^{1/2}\right)^{x - 6} < 5^{-1}$$ Simplifying the exponent: $$5^{\frac{x - 6}{2}} < 5^{-1}$$ Equating the exponents: $$\frac{x - 6}{2} < -1 \quad \Rightarrow \quad x - 6 < -2 \quad \Rightarrow \quad x < 4$$ 2) $$\left(\frac{2}{13}\right)^{x^2 - 1} \geq 1$$ Since any base raised to 0 equals 1, we solve: $$x^2 - 1 \geq 0 \quad \Rightarrow \quad x^2 \geq 1 \quad \Rightarrow \quad x \geq 1 \quad \text{or} \quad x \leq -1$$ ### 5. Solve the equation: $$7^{x+1} + 3 \cdot 7^x = 2x^5 + 3 \cdot 2x$$ This equation seems to involve complex exponential and polynomial terms, and would require further methods like substitution or numerical solving depending on the given terms. --- Let me know if you'd like more details on any specific step, or if you have any questions! Here are 5 related problems: 1. Solve the equation $$\left(\frac{1}{4}\right)^{x-2} = 16$$. 2. Solve the inequality $$\left(\frac{5}{6}\right)^x < \frac{2}{5}$$. 3. Solve the system of equations $$\begin{cases} 2x + 3y = 7 \\ 4x - y = 5 \end{cases}$$. 4. Solve the inequality $$\left(2^{x-3}\right) > \frac{1}{8}$$. 5. Solve the equation $$5^{x+2} + 25^{x-1} = 30$$. **Tip:** When solving systems of equations, elimination or substitution are both powerful methods! Choose the one that simplifies the problem faster.

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Math Problem Analysis

Mathematical Concepts

Algebra
Exponential equations
Quadratic equations
Inequalities
System of linear equations

Formulas

Exponential form transformation
Quadratic formula: ax^2 + bx + c = 0
Logarithmic properties
Linear equation solving methods

Theorems

Properties of exponents
Quadratic equation solutions
Logarithm rules

Suitable Grade Level

Grades 9-11