log-2-x-2-7x-2-log-2-x-2-3x-6-log-4-8-find-x-

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}

\dots \dots . .

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!