How do you solve $32^(5p+2)>=16^(5p)$ ?

Question

1686 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-10T20:11:25

$p ≥ -2$

Rewrite in powers of $2$ .

$(2^5)^(5p + 2) ≥ (2^4)^(5p)$

$2^(25p + 10) ≥ 2^(20p)$

$25p + 10 ≥ 20p$

$5p ≥ -10$

$p ≥ -2$

Hopefully this helps!

How do you solve 32^(5p+2)>=16^(5p)32^(5p+2)>=16^(5p)?