Math Problem Statement

Log2x(x-1)+logx=log12

Solution

We are tasked with solving the equation:

log2x(x1)+logx=log12\log_2 x (x - 1) + \log x = \log 12

First, let's break down the equation and apply logarithmic properties. Here's how we approach it:

Step 1: Simplify the first logarithmic term

The equation is:

log2x(x1)+logx=log12\log_2 x (x - 1) + \log x = \log 12

But there's an ambiguity in the term log2x(x1)\log_2 x (x - 1). It could either be:

  1. log2(x(x1))\log_2 (x(x - 1))
  2. Or log2x(x1)\log_2 x \cdot (x - 1)

To proceed correctly, can you confirm which interpretation you'd like?

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Math Problem Analysis

Mathematical Concepts

Logarithms
Properties of Logarithms
Logarithmic Equations

Formulas

Logarithmic properties: log_b(xy) = log_b(x) + log_b(y)
Change of base formula: log_b(x) = log_c(x) / log_c(b)
Simplification of logarithmic terms

Theorems

Properties of Logarithms
Logarithmic Equations

Suitable Grade Level

Grades 10-12