Menu Close

5-x-1-2-625-5-x-2-2-5-2x-3-




Question Number 115246 by bemath last updated on 24/Sep/20
    5^((x+1)^2 )  + 625 ≤ 5^(x^2 +2)  + 5^(2x+3)
$$\:\:\:\:\mathrm{5}^{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \:+\:\mathrm{625}\:\leqslant\:\mathrm{5}^{{x}^{\mathrm{2}} +\mathrm{2}} \:+\:\mathrm{5}^{\mathrm{2}{x}+\mathrm{3}} \\ $$
Commented by bobhans last updated on 24/Sep/20
let 5^x^2   = r ∧ 5^(2x)  = s
$${let}\:\mathrm{5}^{{x}^{\mathrm{2}} } \:=\:{r}\:\wedge\:\mathrm{5}^{\mathrm{2}{x}} \:=\:{s} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *