Find all values of x which satisfy the inequality

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Homework Help Overview

The discussion revolves around solving the inequality involving exponential expressions: 6^(2x+3) < 2^(x+7) * 3^(3x-1). Participants are exploring methods to manipulate and analyze the inequality.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to simplify the inequality but expresses uncertainty about the correctness of their approach. Some participants question the intent behind the inequality, asking whether the goal is to find specific values of x, demonstrate the inequality's validity, or prove it false. Others suggest taking logarithms as a potential method for solving the inequality.

Discussion Status

Participants are actively engaging with the problem, with some providing hints and alternative approaches. There is a mix of strategies being discussed, and while no consensus has been reached, the conversation is progressing with various interpretations being explored.

Contextual Notes

There is a mention of the original poster's oversight in clearly stating their objective, which may affect the direction of the discussion. Additionally, some participants note that the numbers involved in the inequality might simplify the problem in unexpected ways.

Petkovsky
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6^(2x+3) < 2^(x+7) * 3^(3x-1)

So what i did first was:

3^(2x+3) * 2^(2x+3) < 2^(x+7) * 3^(3x-1)

Now i don't know how to set up the equation. I guess this is not correct

2*(2x+3) < x+7 + 3x - 1
 
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First of all, what is it that you're trying to do? Find for which x this is true? Show that it is always true? Show that it is never true?
 
Find all values of x which satisfy the inequality. Sorry i didnt mention, i thought it was clear.
 
Petkovsky said:
6^(2x+3) < 2^(x+7) * 3^(3x-1)

Hi Petkovsky! :smile:

Hint: take logs. :wink:
 
We needn't even take logs, the numbers happen to work out very nicely =]

Q: Find x such that; [tex]6^{2x+3} < 2^{x+7} \cdot 3^{3x-1}[/tex].

Rewrite the exponents on the RHS to also have 2x+3's, [tex]RHS = \frac{2^{2x+3}}{2^{x-4}} \cdot 3^{2x+3} 3^{x-4} = 6^{2x+3} \left( \frac{3}{2} \right)^{ x-4}[/tex].

The question is much easier in this form.
 

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