Sigma Percentile
JEE Advanced 2022
LEVELJEE Main

Animated Solution for Mathematics - Basic Mathematics: The product of all positive real values of satisfying the equation is ______.

Enter Numerical Value:

Visualized Solution

Analyze the Equation

  • Given equation:
  • Notice the exponent contains logarithms with base .
  • We need to find the product of all positive real values of .

Taking Logarithm on Both Sides

  • To simplify the complex exponent, we take logarithm on both sides.
  • Since the logarithms in the exponent have base , we apply .

Applying Log Properties

  • Use the power rule of logarithms:
  • The entire exponent comes down as a multiplier.

Simplifying the RHS

  • Evaluate the right-hand side:
  • Using the property , we get .

Introducing Substitution

  • The equation looks complicated with everywhere.
  • Let's use a substitution to make it a simple polynomial.
  • Let

Substituting

  • Replace every with .

Expanding the Equation

  • Multiply inside the bracket.
  • Rearrange to form a standard polynomial equation:

Simplifying the Polynomial

  • Notice that all coefficients are multiples of .
  • Divide the entire equation by to simplify calculations.

Recognizing the Quadratic Form

  • This is a bi-quadratic equation in .
  • It can be treated as a quadratic equation in terms of .
  • Let

Equation in

  • Substitute into the simplified equation.

Factorizing the Quadratic

  • We need two numbers that multiply to and add to .
  • These numbers are and .
  • Split the middle term:

Solving for

  • Set each factor to zero.

Solving for

  • Recall that .
  • Case 1:
  • Case 2:
  • So, can be .

Solving for

  • Recall our initial substitution:
  • For ,
  • For ,
  • For ,
  • For ,

Final Calculation

  • We need the product of all these values of .
  • Using exponent rules: add the powers.
  • Key Takeaway: Logarithmic substitutions can transform complex exponential equations into solvable polynomials.

The Sigma Insight: Logarithmic Equations and Inequalities

Analyzing the Setup

We are presented with the equation:
This problem features a variable in both the base and the exponent, combined with logarithmic terms. To simplify this, we apply the logarithmic lens by taking on both sides.

The Power of Transformation

Applying to both sides and utilizing the power rule , we bring the exponent down:
Using the property , the right-hand side simplifies to . Our equation now becomes:

The Great Simplifier

To manage the complexity, we introduce the substitution . This transforms the equation into a polynomial:
Distributing yields the bi-quadratic equation:
Letting , we simplify the expression to . Dividing the entire equation by , we obtain:

The Final Reveal

We factorize the quadratic by splitting the middle term:
This yields two possible values for : or . Since , we solve for : 1. If , then . 2. If , then .
Translating back to using , the four possible values for are , , , and . The product of these values is:
The final product of all possible values of is 1.

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