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

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

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Answer 1 · 2018-03-10T20:11:25

$p \geq - 2$ $p ≥ -2$

Explanation:

Rewrite in powers of $2$ $2$ .

${(2^{5})}^{5 p + 2} \geq {(2^{4})}^{5 p}$ $(2^5)^(5p + 2) ≥ (2^4)^(5p)$

$2^{25 p + 10} \geq 2^{20 p}$ $2^(25p + 10) ≥ 2^(20p)$

$25 p + 10 \geq 20 p$ $25p + 10 ≥ 20p$

$5 p \geq - 10$ $5p ≥ -10$

$p \geq - 2$ $p ≥ -2$

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

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Explanation:

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