I have a different way: \(\displaystyle a^2-5a+6<0\) or \(\displaystyle (a-2)(a-3)<0\),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 \(\displaystyle a=7^x\).
One Comment: It was posted in the Pre-Algebra Forum.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?