Find all values of x which satisfy the inequality

[Petkovsky]

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

 

[nicksauce]

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?

 

[Petkovsky]

Find all values of x which satisfy the inequality. Sorry i didnt mention, i thought it was clear.

 

[tiny-tim]

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

Hi Petkovsky!

Hint: take logs.

 

[Gib Z]

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

Q: Find x such that; 6^{2x+3} &lt; 2^{x+7} \cdot 3^{3x-1}.

Rewrite the exponents on the RHS to also have 2x+3's, 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}.

The question is much easier in this form.