How do you solve the exponential inequality $2^{x + 3} \leq 16^{4 x - 2}$ $2^(x+3)<=16^(4x-2)$ ?

Question

1312 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-17T10:18:46

$x \geq \frac{11}{15}$ $x ge 11/15$

Knowing that $16 = 2^{4}$ $16=2^4$ we have

$2^{x + 3} \leq 2^{4 (4 x - 2)}$ $2^(x+3) le 2^(4(4x-2))$

Here $2 > 1$ $2 > 1$ then if $2^{α} \leq 2^{β} \to α \leq β$ $2^alpha le 2^beta->alpha le beta$

so

$x + 3 \leq 16 x - 8 \to 15 x \geq 11 \to x \geq \frac{11}{15}$ $x+3 le 16x-8->15xge 11->x ge 11/15$

How do you solve the exponential inequality 2x+3≤164x−22^(x+3)<=16^(4x-2)?