Math Problem Statement

log_x 1000 − log_x x^2 + (10^-3)^0 = 0

Solution

The equation in the image is:

logx1000logxx2+(103)0=0\log_x 1000 - \log_x x^2 + (10^{-3})^0 = 0

Let's break it down step by step:

Step 1: Simplify (103)0(10^{-3})^0

Any non-zero number raised to the power of 0 equals 1, so: (103)0=1(10^{-3})^0 = 1 Thus, the equation becomes: logx1000logxx2+1=0\log_x 1000 - \log_x x^2 + 1 = 0

Step 2: Apply logarithmic rules

Using the logarithmic identity logxxn=n\log_x x^n = n, we can simplify logxx2\log_x x^2: logxx2=2\log_x x^2 = 2 Now the equation becomes: logx10002+1=0\log_x 1000 - 2 + 1 = 0

Step 3: Combine constants

Simplify the constants: logx10001=0\log_x 1000 - 1 = 0 This simplifies further to: logx1000=1\log_x 1000 = 1

Step 4: Solve for xx

The equation logx1000=1\log_x 1000 = 1 means that xx raised to the power of 1 equals 1000: x1=1000x^1 = 1000 Thus, x=1000x = 1000.

Final Answer:

The value of xx is 10001000.

Let me know if you'd like further clarification or additional details.


Here are 5 related questions to expand this concept:

  1. What is the general rule for solving logarithmic equations?
  2. How do we handle logarithmic expressions when the base of the logarithm is unknown?
  3. Can logarithms have negative bases?
  4. What happens when the logarithmic base is a fraction between 0 and 1?
  5. How do you convert between logarithmic and exponential forms?

Tip: Always simplify any constants or expressions before applying logarithmic rules to avoid mistakes.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Exponents
Algebra

Formulas

log_x a − log_x b = log_x (a / b)
log_x x^n = n
a^0 = 1

Theorems

Properties of Logarithms
Exponentiation Rules

Suitable Grade Level

Grades 9-12