How Do You Solve Complex Exponential Inequalities?

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The discussion focuses on solving complex exponential inequalities, specifically three problems involving exponential expressions. The second inequality is addressed with a step-by-step simplification, but the user encounters an error when substituting values. A suggested approach involves rewriting the inequality in terms of a new variable, u, to simplify the expression. The transformation leads to a quadratic inequality that can be solved more easily. Clarification on the simplification process and solving methods for exponential inequalities is requested for further assistance.
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Homework Statement


1) 5x+4-5x-1<24
2) 2x+2-1>=22x+3
3) 21-x-2x+1/2x-1<0

Homework Equations


At the 2nd I do like this:
2x+2-22x>=4 divide by 2
2x+1-22x-1>=2
x+1-2x+1>=1
-x>=-1 divide by (-1)
x=<1
but when i put 0 in x the inequation is incorrect, what I have mistake, please tell me.
If you can help me in anothers inequations, thanks a lot.

The Attempt at a Solution

 
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Please can you help for this?
 
I need the help urgently, please help!
 
Your simplifications with the exponents aren't correct. Try this:

<br /> \begin{align*}<br /> 2^{x+2} - 2^{2x} &amp; \ge 4\\<br /> 2^x \cdot 2^2 - 2^{2x} &amp; \ge 4\\<br /> 4\cdot 2^x - 2^{2x} -4 &amp; \ge 0\\<br /> 2^{2x} - 4\cdot 2^x + 4 &amp; \le 0<br /> \end{align*}<br />

Now, for ease of reading, let

<br /> u = 2^x \Rightarrow u^2 = 2^{2x}<br />

so the inequality is now

<br /> u^2 - 4u + 4 &amp; \le 0<br />
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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