3x-1-log-3-3x-gt-81x-2-find-the-solution-set-

Question Number 105843 by bobhans last updated on 01/Aug/20

{(3 x)}^{1 + \log_{3} (3 x)} > 81 x^{2}

f i n d t h e s o l u t i o n s e t

Answered by bemath last updated on 01/Aug/20

{(3 x)}^{1 + \log_{3} (3 x)} > 9 {(3 x)}^{2}

(1) d o m a i n x > 0

(2) (1 + \log_{3} (3 x)) (\log_{3} (3 x)) >

\log_{3} (9 . {(3 x)}^{2})

\Rightarrow l e t \log_{3} (3 x) = b

(1 + b) (b) > 2 + 2 b

b^{2} + b - 2 b - 2 > 0

b^{2} - b - 2 > 0 \to {\begin{cases} b < - 1 \\ b > 2 \end{cases}

\to {\begin{cases} \log_{3} (3 x) < \log_{3} (\frac{1}{3}) \\ \log_{3} (3 x) > \log_{3} (9) \end{cases}

\to {\begin{cases} x < \frac{1}{9} \\ x > 3 \end{cases}

s o l u t i o n s e t : 0 < x < \frac{1}{9} \cup x > 3

Commented by Coronavirus last updated on 01/Aug/20

clean

Answered by bobhans last updated on 01/Aug/20

{(3 x)}^{1 + \log_{3} (3 x)} > {(3 x)}^{\log_{(3 x)} (81 x^{2})}

(3 x - 1) (1 + \log_{3} (3 x) - \log_{(3 x)} (81 x^{2})) > 0

(3 x - 1) (2 + \log_{3} (x) - {\frac{4 + 2 \log_{3} (x)}{1 + \log_{3} (x)}}) > 0

(3 x - 1) (2 + \log_{3} (x)) (1 - \frac{2}{1 + \log_{3} (x)}) > 0

(3 x - 1) (2 + \log_{3} (x)) (\frac{\log_{3} (x) - 1}{1 + \log_{3} (x)}) > 0

(1) s i n c e x > 0 t h e n 3 x - 1 > 0

s o \frac{(2 + \log_{3} (x)) (\log_{3} (x) - 1)}{1 + \log_{3} (x)} > 0

l e t \log_{3} (x) = υ \Rightarrow \frac{(2 + ϑ) (ϑ - 1)}{ϑ + 1} > 0

\to {\begin{cases} - 2 < ϑ < - 1 \to \frac{1}{9} < x < \frac{1}{3} \\ ϑ > 1 \to x > 3 \end{cases}

Answered by 1549442205PVT last updated on 01/Aug/20

We need the condition x > 0

Take logarithm both two side of

ineqn . by base 3 we get

(1 + \log_{3} (3 x)) \log_{3} (3 x) > \log_{3} ({81 x}^{2}) = \log_{3} 81 + \log_{3} (x^{2})

\Leftrightarrow (2 + \log_{3} x) (1 + \log_{3} x) > {2 \log}_{3} x + 4

Set \log_{3} x = y we get

(y + 2) (1 + y) - 2 y - 4 > 0

\Leftrightarrow y^{2} + y - 2 > 0 \Leftrightarrow (y - 1) (y + 2) > 0

\Leftrightarrow [\begin{matrix} y > 1 \\ y < - 2 \end{matrix}] \Leftrightarrow [\begin{matrix} \log_{3} x > 1 \Leftrightarrow x > 3 \\ \log_{3} x < - 2 \Leftrightarrow 0 < x < \frac{1}{9} \end{matrix}]

Thus, the roots of the given ineqn . is

x \in (0; \frac{1}{9}) \cup (3; + \infty)