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Question Number 163226 by nurtani last updated on 05/Jan/22

Commented by Rasheed.Sindhi last updated on 06/Jan/22

$Sir mr W, w h y h a v e y o u d e l e t e d y o u r$ $a n s w e r ?$
Commented by mr W last updated on 06/Jan/22

$s o m e t i m e s i d e l e t e m y a n s w e r w h e n$ $i s e e t h a t t h e r e e x i s t s a l r e a d y t h e$ $s a m e o r s i m i l a r a n s w e r f r o m o t h e r$ $p e o p l e .$
Commented by Rasheed.Sindhi last updated on 06/Jan/22

$U n d e r s t o o d s i r!$
	Answered by cortano1 last updated on 05/Jan/22

	$2^{5 x - 1} . 3^{4 x + 1} . 7^{3 x + 3} = 2^{3 x - 6} . 3^{2 x - 4} . 7^{x - 2}$ $\Rightarrow x = - \frac{5}{2}$
	Answered by Rasheed.Sindhi last updated on 05/Jan/22

	$2^{5 x - 1} . 3^{4 x + 1} . 7^{3 x + 3} = {(2^{3} . 3^{2} . 7)}^{x - 2}$ $= 2^{3 x - 6} . 3^{2 x - 4} . 7^{x - 2)}$ $2^{5 x - 1 - 3 x + 6} . 3^{4 x + 1 - 2 x + 4} . 7^{3 x + 3 - x + 2} = 1$ $2^{2 x + 5} . 3^{2 x + 5} . 7^{2 x + 5} = 1$ ${(2.3.7)}^{2 x + 5} = {(2.3.7)}^{0}$ $2 x + 5 = 0$ $x = - \frac{5}{2}$
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