log_2 (x^2 +7x−2)=log_2 (x^2 +3x−6)+log


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Question Number 55224 by Otchere Abdullai last updated on 19/Feb/19

${l o g}_{2} (x^{2} + 7 x - 2) = {l o g}_{2} (x^{2} + 3 x - 6) + {l o g}_{4} 8$ $f i n d x$
	Answered by peter frank last updated on 19/Feb/19

	$\log_{2} (\frac{x^{2} + 7 x - 2}{x^{2} + 3 x - 6}) = {3 \log}_{2^{2}} 2$ $\log_{2} (\frac{x^{2} + 7 x - 2}{x^{2} + 3 x - 6}) = \frac{\log 8}{\log 4}$ $\log_{2} (\frac{x^{2} + 7 x - 2}{x^{2} + 3 x - 6}) = \frac{3 \log 2}{2 \log 2}$ $\log_{2} (\frac{x^{2} + 7 x - 2}{x^{2} + 3 x - 6}) = \frac{3}{2}$ $2^{\frac{3}{2}} = \frac{x^{2} + 7 x - 2}{x^{2} + 3 x - 6}$ $. . . . . . . .$
	Commented by Otchere Abdullai last updated on 19/Feb/19

	$t h a n k s s i r b u t i n t h e b o o k t h e a n s w e r$ $i s 7 b u t w a s n o t s o l v e d$
	Answered by kaivan.ahmadi last updated on 19/Feb/19

	${l o g}_{2} (\frac{x^{2} + 7 x - 2}{x^{2} + 3 x - 6}) = {l o g}_{2^{2}} 8 = \frac{1}{2} {l o g}_{2} 8 = {l o g}_{2} \sqrt{8} \Rightarrow$ $x^{2} + 7 x - 2 = \sqrt{8} x^{2} + 3 \sqrt{8} x - 6 \sqrt{8} \Rightarrow$ $(\sqrt{8} - 1) x^{2} + (3 \sqrt{8} - 7) x - (6 \sqrt{8} - 2) = 0$ $Δ = {(3 \sqrt{8} - 7)}^{2} - 4 (\sqrt{8} - 1) (6 \sqrt{8} - 2) =$ $72 - 42 \sqrt{8} + 49 - 4 (48 - 8 \sqrt{8} + 2) =$ $121 - 42 \sqrt{8} - 192 + 32 \sqrt{8} - 8 =$ $- 79 - 10 \sqrt{8} < 0$ $\Rightarrow t h e r e i s n o r e a l n u m b e r f o r x$
	Commented by Otchere Abdullai last updated on 19/Feb/19

	$t h a n k s s i r t h e a n s w e r i n t h e b o o k i s$ $x = 7 b u t u n s o l v e d q u e s t i o n$
	Commented by kaivan.ahmadi last updated on 19/Feb/19

	$b u t i f y o u r e p l a c e x = 7 t h e e q u a l i t y i s n o t t r u e$
	Commented by Otchere Abdullai last updated on 19/Feb/19

	$o k t h a n k s s i r i w i l l r e - c h e c k t h e$ $q u e s t i o n f r o m t h e l i b r a r y$
	Answered by MJS last updated on 20/Feb/19

	$\log_{2} a = \frac{\ln a}{\ln 2}$ $\log_{4} 8 = \frac{\ln 8}{\ln 4} = \frac{3 \ln 2}{2 \ln 2} = \frac{3}{2}$ $\ln (x^{2} + 7 x - 2) = \ln (x^{2} + 3 x - 6) + \frac{3 \ln 2}{2}$ $\ln \frac{x^{2} + 7 x - 2}{x^{2} + 3 x - 6} = \frac{3 \ln 2}{2}$ $\frac{x^{2} + 7 x - 2}{x^{2} + 3 x - 6} = 2 \sqrt{2}$ $transforming$ $x^{2} + \frac{17 - 8 \sqrt{2}}{7} x - \frac{2 (23 + 4 \sqrt{2})}{7} = 0$ $x = - \frac{17 - 8 \sqrt{2}}{14} \pm \frac{\sqrt{1705 - 48 \sqrt{2}}}{14}$ $both satisfy the given equation although$ $the ‘ ‘ -^{″} solution leads to complex values$ $for both logarithms$
	Commented by Otchere Abdullai last updated on 20/Feb/19

	$T h a n k y o u m j s s i r!$
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