Math Problem Statement

ដោះស្រាយសមីការ អិចស្ប៉ូណង់ស្យែល : a. 4^(x-1) = (1/2)^(1-3x) B. 4^(x) × 8^(2x-1) = 1/16 C. 4^x - 3.2^x + 2 = 0 D. 4.9^x - 13.6^x + 9.4^x = 0 E. 27^x + 12^x - 2.8^x = 0 Please tell me

Solution

Let's solve each exponential equation step by step.

A. Solve 4x1=(12)13x4^{x-1} = \left(\frac{1}{2}\right)^{1-3x}

  1. Rewrite both sides with the same base:

    4x1=(21)13x4^{x-1} = \left(2^{-1}\right)^{1-3x}

    4x1=21(13x)=2(13x)=23x14^{x-1} = 2^{-1(1-3x)} = 2^{-(1-3x)} = 2^{3x-1}

    Since 4=224 = 2^2, rewrite the left side:

    (22)x1=22(x1)=22x2(2^2)^{x-1} = 2^{2(x-1)} = 2^{2x-2}

    Now we have:

    22x2=23x12^{2x-2} = 2^{3x-1}

  2. Set the exponents equal:

    2x2=3x12x - 2 = 3x - 1

  3. Solve for xx:

    2x3x=1+22x - 3x = -1 + 2

    x=1-x = 1

    x=1x = -1

B. Solve 4x×82x1=1164^x \times 8^{2x-1} = \frac{1}{16}

  1. Rewrite both sides with the same base:

    4x×82x1=424^x \times 8^{2x-1} = 4^{-2}

    4=224 = 2^2 and 8=238 = 2^3, so:

    (22)x×(23)2x1=24(2^2)^x \times (2^3)^{2x-1} = 2^{-4}

    22x×26x3=242^{2x} \times 2^{6x-3} = 2^{-4}

    Combine exponents:

    22x+6x3=242^{2x + 6x - 3} = 2^{-4}

    28x3=242^{8x-3} = 2^{-4}

  2. Set the exponents equal:

    8x3=48x - 3 = -4

  3. Solve for xx:

    8x=4+38x = -4 + 3

    8x=18x = -1

    x=18x = -\frac{1}{8}

C. Solve 4x32x+2=04^x - 3 \cdot 2^x + 2 = 0

  1. Substitute y=2xy = 2^x, so 4x=(22)x=y24^x = (2^2)^x = y^2:

    y23y+2=0y^2 - 3y + 2 = 0

  2. Factor the quadratic equation:

    (y1)(y2)=0(y - 1)(y - 2) = 0

    So, y=1y = 1 or y=2y = 2.

  3. Back-substitute y=2xy = 2^x:

    • For y=1y = 1: 2x=12^x = 1 gives x=0x = 0.
    • For y=2y = 2: 2x=22^x = 2 gives x=1x = 1.

    So, x=0x = 0 or x=1x = 1.

D. Solve 4.9x13.6x+9.4x=04.9^x - 13.6^x + 9.4^x = 0

This equation is nonlinear and cannot be solved using simple algebraic methods. Numerical methods, such as graphing or using a calculator, are needed to approximate the solution. For now, we leave this unsolved analytically.

E. Solve 27x+12x2.8x=027^x + 12^x - 2.8^x = 0

Similarly, this equation involves different bases and exponents, making it difficult to solve algebraically. Numerical methods or a calculator are required to find the solution. This also will be left unsolved analytically.

Summary of Solutions:

  • A. x=1x = -1
  • B. x=18x = -\frac{1}{8}
  • C. x=0x = 0 or x=1x = 1
  • D. Requires numerical methods
  • E. Requires numerical methods

Would you like further details or have any questions? Here are 5 related questions to consider:

  1. How do you solve exponential equations when the bases are different?
  2. What methods can be used to solve exponential equations numerically?
  3. How do logarithms help in solving exponential equations?
  4. What are some common properties of exponents used in these solutions?
  5. How can quadratic equations be used in solving exponential problems?

Tip: Always try to express both sides of an exponential equation with the same base to simplify the equation.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Exponential Equations
Logarithms
Quadratic Equations

Formulas

Exponential equation rules
Logarithm rules
Quadratic equation formula

Theorems

-

Suitable Grade Level

Grades 10-12