log-a-3x-4a-log-a-3x-2-log-2-a-log-a-1-2a-where-0-lt-a-lt-1-2-find-the-value-of-x-

Question Number 109775 by ZiYangLee last updated on 25/Aug/20

\log_{a} (3 x - 4 a) + \log_{a} 3 x = \frac{2}{\log_{2} a} + \log_{a} (1 - 2 a), where 0 < a < \frac{1}{2},

find the value of x .

Answered by bemath last updated on 25/Aug/20

△ \frac{♭ e}{m a t h} ▽

\log_{a} (3 x (3 x - 4 a)) = 2 \log_{a} (2) + \log_{a} (1 - 2 a)

\log_{a} (9 x^{2} - 12 a x) = \log_{a} (4 (1 - 2 a))

9 x^{2} - 12 a x = 4 - 8 a

9 x^{2} - 12 a x + 8 a - 4 = 0; w h e r e x > \frac{4 a}{3}, 0 < a < \frac{1}{2}

x = \frac{12 a + \sqrt{144 a^{2} - 4.9 (8 a - 4)}}{18}

x = \frac{12 a + 12 \sqrt{a^{2} - (2 a - 1)}}{18}

x = \frac{2 a + 2 \sqrt{a^{2} - 2 a + 1}}{3} = \frac{2 a + 2 \sqrt{{(a - 1)}^{2}}}{3}

x = \frac{2 a + 2 ∣ a - 1 ∣}{3} = \frac{2 a - 2 (a - 1)}{3} = \frac{2}{3} \leftarrow t h e a n s w e r

n o t e ∣ a - 1 ∣ = - (a - 1) f o r 0 < a < \frac{1}{2}

Commented by ZiYangLee last updated on 25/Aug/20

Why \log_{a} (4 a (1 - 2 a)) at the second row ?

Commented by bemath last updated on 25/Aug/20

\frac{1}{\log_{2} (a)} = \log_{a} (2)

Commented by Coronavirus last updated on 25/Aug/20

\log_{2} (a) = \frac{\ln (a)}{\ln (2)} \Rightarrow \frac{2}{\log_{2} (a)} = \frac{2 \ln (2)}{\ln (a)} = \frac{\ln (4)}{\ln (a)} = \log_{a} (4)

Answered by Rasheed.Sindhi last updated on 25/Aug/20

\log_{a} (3 x - 4 a) + \log_{a} 3 x - \log_{a} (1 - 2 a)

= \frac{2}{\log_{2} a}

\log_{a} (\frac{3 x (3 x - 4 a)}{(1 - 2 a)}) = \frac{2}{\frac{1}{\log_{a} 2}}

\log_{a} (\frac{3 x (3 x - 4 a)}{(1 - 2 a)}) = {2 \log}_{a} 2

\log_{a} (\frac{3 x (3 x - 4 a)}{(1 - 2 a)}) = \log_{a} 2^{2}

\frac{3 x (3 x - 4 a)}{(1 - 2 a)} = 2^{2}

9 x^{2} - 12 a x = 4 (1 - 2 a)

9 x^{2} - 12 a x - 4 (1 - 2 a) = 0

x = \frac{12 a \pm \sqrt{144 a^{2} - 4 (9) (- 4 (1 - 2 a)}}{2 (9)}

x = \frac{12 a \pm \sqrt{144 a^{2} + 144 (1 - 2 a)}}{18}

x = \frac{12 a \pm 12 \sqrt{a^{2} - 2 a + 1}}{18}

x = \frac{2 a \pm 2 (a - 1)}{3} = \frac{2 a \pm 2 a \mp 2}{3}

x = 4 a - 2, 2 ◂

Answered by Her_Majesty last updated on 25/Aug/20

a > 0 \land a \neq 1 \land 1 - 2 a > 0 \land 1 - 2 a \neq 1 \Leftrightarrow 0 < a < \frac{1}{2}

3 x > 0 \land 3 x - 4 a > 0 \Leftrightarrow x > \frac{4 a}{3}

\frac{l n (3 x - 4 a)}{l n a} + \frac{l n 3 x}{l n a} = \frac{2 l n 2}{l n a} + \frac{l n (1 - 2 a)}{l n a}

l n (3 x (3 x - 4 a)) = l n (4 (1 - 2 a))

3 x (3 x - 4 a) = 4 (1 - 2 a)

x^{2} - \frac{4 a}{3} x + \frac{4 (2 a - 1)}{9} = 0

(x - \frac{2}{3}) (x - \frac{2 (2 a - 1)}{3}) = 0

x_{1} = \frac{2}{3}

x_{2} = \frac{2 (2 a - 1)}{3} b u t 0 < a < \frac{1}{2} \land x > \frac{4 a}{3} m a k e s

t h i s i m p o s s i b l e :

x > \frac{4 a}{3} \land x = \frac{4 a}{3} - \frac{1}{3} i s f a l s e