3^x .8^(x/(x+2)) =6ln0−log10


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Question Number 187406 by Fridunatjan08 last updated on 16/Feb/23

$3^{x} . 8^{\frac{x}{x + 2}} = 6 l n 0 - l o g 10$
Commented by Fridunatjan08 last updated on 16/Feb/23

$c a n s o m e o n e p l e a s e h e l p m e ?$
Commented by Frix last updated on 17/Feb/23

$\ln 0 is not defined$
	Answered by a.lgnaoui last updated on 17/Feb/23

	$I t h i n k t h a t y o u w a n t t o t e l 6 l n 10$ $w e s u p p o s e t h a t t h r u e$ $l o g 10 = \frac{l n 10}{l n 10} = 1$ $3^{x} . 8^{\frac{x}{x + 2}} = 6 l n 10 - 1$ $a = b (a > 0 b > 0 \Rightarrow l n a = l n b$ $x l n 3 + \frac{x}{x + 2} l n 8 = l n [6 l n 10 - 1]$ $p o s o n s k = l n [6 l n 10 - 1]$ $x (x + 2) l n 3 + x l n 8 = (x + 2) k$ $x^{2} l n 3 + 2 x l n 3 + x l n 8 = (x + 2) k$ $x^{2} l n 3 + x (2 l n 3 + l n 8 - k) - 2 k = 0$ $x^{2} + \frac{1}{l n 3} (2 l n 3 + 3 l n 2 - k) x - \frac{2 k}{l n 3} = 0$ ${[x + (1 + \frac{3 l n 2 - k}{2 l n 3})]}^{2} -$ ${[1 + \frac{3 l n 2 - k}{2 l n 3})}^{2} + \frac{2 k}{l n 3}] = 0$ $x + (1 + \frac{3 l n 2 - k}{2 l n 3}) =$ $\sqrt{{(1 + \frac{3 l n 2 - k}{2 l n 3})}^{2} + \frac{2 k}{l n 3}}$ $x = \pm \sqrt{{(1 + \frac{3 l n 2 - k}{2 l n 3})}^{2} + \frac{2 k}{l n 3}} - 1 - \frac{3 k n 2 - k}{2 l n 3}$
	Commented by Fridunatjan08 last updated on 17/Feb/23

	$t h s n k s y o u b r o .$
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