inequality 7^(2x)-5*7^(x)+6<0

acemi123

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Apr 14, 2019
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I couldn't figure out to solve this inequality. Any help pls.

7^(2x)-5*7^(x)+6<0
 
What if we rewrite the inequality to read:

[MATH](7^x)^2-5\cdot7^x+6<0[/MATH]
Do you now see the LHS is a quadratic in \(7^x\)? Can you proceed?
 
I couldn't figure out to solve this inequality. Any help pls.
7^(2x)-5*7^(x)+6<0
I have a different way: \(\displaystyle a^2-5a+6<0\) or \(\displaystyle (a-2)(a-3)<0\),
What areas satisfy that inequality: \(\displaystyle (-\infty,2),~(2,3),\text{ or }(3,\infty)~?\)
OH, let \(\displaystyle a=7^x\).
 
I have a different way: \(\displaystyle a^2-5a+6<0\) or \(\displaystyle (a-2)(a-3)<0\),
What areas satisfy that inequality: \(\displaystyle (-\infty,2),~(2,3),\text{ or }(3,\infty)~?\)
OH, let \(\displaystyle a=7^x\).

That was exactly where I was hoping to lead the OP. Oh well.
 
What if we rewrite the inequality to read:

[MATH](7^x)^2-5\cdot7^x+6<0[/MATH]
Do you now see the LHS is a quadratic in \(7^x\)? Can you proceed?

If we factor, we find:

[MATH](7^x-2)(7^x-3)<0[/MATH]
From which we find:

[MATH]2<7^x<3[/MATH]
Hence:

[MATH]\log_7(2)<x<\log_7(3)[/MATH]
Here is a plot of the curve, showing the interval for which it is negative:

fmh_0056.png
 
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