Math Problem Statement

Simplify the mathematical expression: \[ \frac{1 - \log_a^3 b}{(\log_a b + \log_b a + 1) \log_a \frac{a}{b}} \]

Solution

The given mathematical expression is:

1loga3b(logab+logba+1)logaab\frac{1 - \log_a^3 b}{\left( \log_a b + \log_b a + 1 \right) \log_a \frac{a}{b}}

Let's simplify it step by step.


Step 1: Simplify loga3b\log_a^3 b

The term loga3b\log_a^3 b is simply shorthand for (logab)3(\log_a b)^3.


Step 2: Simplify logaab\log_a \frac{a}{b}

Using the logarithmic property:

logaab=logaalogab=1logab\log_a \frac{a}{b} = \log_a a - \log_a b = 1 - \log_a b

So the denominator becomes:

(logab+logba+1)(1logab)\left( \log_a b + \log_b a + 1 \right)(1 - \log_a b)


Step 3: Simplify logba\log_b a

Using the base change property of logarithms:

logba=1logab\log_b a = \frac{1}{\log_a b}

Substituting this into the denominator:

(logab+1logab+1)(1logab)\left( \log_a b + \frac{1}{\log_a b} + 1 \right)(1 - \log_a b)


Step 4: Analyze the full expression

Now the expression becomes:

1(logab)3(logab+1logab+1)(1logab)\frac{1 - (\log_a b)^3}{\left( \log_a b + \frac{1}{\log_a b} + 1 \right)(1 - \log_a b)}

Step 5: Factor out 1logab1 - \log_a b if possible

The numerator 1(logab)31 - (\log_a b)^3 can be factored using the difference of cubes formula:

1(logab)3=(1logab)(1+logab+(logab)2)1 - (\log_a b)^3 = \left(1 - \log_a b \right)\left(1 + \log_a b + (\log_a b)^2\right)

So the entire expression becomes:

(1logab)(1+logab+(logab)2)(logab+1logab+1)(1logab)\frac{\left(1 - \log_a b\right)\left(1 + \log_a b + (\log_a b)^2\right)}{\left(\log_a b + \frac{1}{\log_a b} + 1\right)\left(1 - \log_a b\right)}


Step 6: Cancel 1logab1 - \log_a b

Assuming 1logab01 - \log_a b \neq 0, we can cancel 1logab1 - \log_a b from the numerator and denominator:

1+logab+(logab)2logab+1logab+1\frac{1 + \log_a b + (\log_a b)^2}{\log_a b + \frac{1}{\log_a b} + 1}


Final Simplified Expression:

1+logab+(logab)2logab+1logab+1\frac{1 + \log_a b + (\log_a b)^2}{\log_a b + \frac{1}{\log_a b} + 1}


Would you like further details on any of these steps, or help applying it to specific values?


Related Questions:

  1. What are the logarithmic properties used in this simplification?
  2. How does the base change formula work in logarithms?
  3. What is the difference of cubes formula, and how is it applied here?
  4. Can this expression be simplified further for specific values of aa and bb?
  5. How does the cancellation process work when terms in the numerator and denominator are similar?

Tip:

When simplifying logarithmic expressions, always look for common properties like the base change formula, product rule, and difference of cubes to make the problem manageable.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Simplification
Difference of Cubes
Base Change Formula

Formulas

Difference of cubes: x^3 - y^3 = (x - y)(x^2 + xy + y^2)
Logarithmic base change formula: \log_b a = \frac{1}{\log_a b}
Logarithmic subtraction: \log_a \frac{x}{y} = \log_a x - \log_a y

Theorems

Logarithmic Properties

Suitable Grade Level

Grade 11-12