How do you solve the exponential inequality $8^{x - 1} \leq 32^{3 x - 2}$ $8^(x-1)<=32^(3x-2)$ ?

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Answer 1 · 2018-06-19T07:28:57

$x \geq \frac{3}{4}$ $x>=3/4$

$8^{x - 1} \leq 32^{3 x - 2}$ $8^(x-1)<=32^(3x-2)$

change the $8 and 32$ $8 " and "32$ to powers of 2

${(2^{3})}^{x - 1} \leq {(2^{5})}^{3 x - 2}$ $(2^3)^(x-1)<=(2^5)^(3x-2)$

ie

$2^{3 x - 1} \leq 2^{15 x - 10}$ $2^(3x-1)<=2^(15x-10)$

$\Rightarrow 3 x - 1 \leq 15 x - 10$ $=>3x-1<=15x-10$

solve for $x$ $x$

$- 1 + 10 \leq 15 x - 3 x$ $-1+10<=15x-3x$

$9 \leq 12 x$ $9<=12x$

$12 x \geq 9$ $12x>=9$

$\Rightarrow x \geq \frac{9}{12}$ $=>x>=9/12$

$x \geq \frac{3}{4}$ $x>=3/4$

How do you solve the exponential inequality 8x−1≤323x−28^(x-1)<=32^(3x-2)?