bobhans-3-x-2-x-1-2-9-x-10-6-x-2-2x-3-find-the-solution-set-

Question Number 106695 by bobhans last updated on 06/Aug/20

You can't use 'macro parameter character #' in math mode

3^{x} - 2^{x + 1} ⩽ \sqrt{2 . 9^{x} - 10 . 6^{x} + 2^{2 x + 3}}

find the solution set

Answered by bemath last updated on 06/Aug/20

\overset{@ bemath @}{}

\frac{3^{x} - 2 . 2^{x}}{2^{x}} ⩽ \frac{\sqrt{2 . 9^{x} - 10 . 6^{x} + 8 . {(2^{x})}^{2}}}{2^{x}}

{(\frac{3}{2})}^{x} - 2 ⩽ \sqrt{2 {(\frac{3}{2})}^{2 x} - 10 {(\frac{3}{2})}^{x} + 8}

set {(\frac{3}{2})}^{x} = t

t - 2 ⩽ \sqrt{2 (t - 1) (t - 4)}

defined on (t - 1) (t - 4) ⩾ 0

\Rightarrow t ⩽ 1 \cup t ⩾ 4 \dots (1)

squaring both sides

t^{2} - 4 t + 4 ⩽ {2 t}^{2} - 10 t + 8

- t^{2} + 6 t - 4 ⩽ 0; t^{2} - 6 t + 4 ⩾ 0

{(t - 3)}^{2} - 5 ⩾ 0

t ⩽ 3 - \sqrt{5} \cup t ⩾ 3 + \sqrt{5} \dots (2)

from (1) \cap (2) we get \to {\begin{cases} t ⩽ 1 \\ t ⩾ 3 + \sqrt{5} \end{cases}

{\begin{cases} {(\frac{3}{2})}^{x} ⩽ 1 \Rightarrow x ⩽ 0 \\ {(\frac{3}{2})}^{x} ⩾ 3 + \sqrt{5} \Rightarrow x ⩾ \log_{(\frac{3}{2})} (3 + \sqrt{5}) \end{cases}

the solution set is x \in (- \infty, 0] \cup

[\log_{(\frac{3}{2})} (3 + \sqrt{5}), \infty)

Commented by bobhans last updated on 06/Aug/20

afdollll

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