sum-of-roots-log-3-x-log-3-2-5-log-x-9-3-log-x-5-

Question Number 198809 by dimentri last updated on 24/Oct/23

s u m o f r o o t s

\log_{3} x + \log_{3} (2, 5) + \log_{x} 9 = 3 + \log_{x} 5

Answered by a.lgnaoui last updated on 24/Oct/23

\frac{logx}{\log 3} + \frac{\log (2, 5)}{\log 3} + \frac{\log 9}{logx} = 3 + \frac{\log 5}{logx}

{(logx)}^{2} + logx . \log (2, 5) + 2 {(\log 3)}^{2}

= 3 \log 3 logx + \log 3 . \log 5

posons logx = z

z^{2} + z [\log (2, 5) - 3 \log 3] + \log 3 (2 \log 3 - \log 5) = 0

△ = {[\log (2, 5) - 3 \log 3]}^{2} - 4 \log 3 (2 \log 3 - \log 5)

\log (2, 5) = \log (\frac{25}{10}) = \log 25 - 1 = 2 \log 5 - 1

z^{2} + z (2 \log 5 - 3 \log 3 - 1) + \log 3 (2 \log 3 - \log 5) = 0

△ = {(2 \log 5 - 3 \log 3 - 1)}^{2} - 8 {(\log 3)}^{2} +

4 \log 3 . \log 5

= 1 + {(\log 3)}^{2} - 16 \log 3 \log 5 + 6 \log 3

+ 4 {(\log 5)}^{2} - 4 \log 5

= {(2 \log 5 - 1)}^{2} + {(\log 3 - 8 \log 5)}^{2}

= + 6 \log 3 - 64 {(\log 5)}^{2}

= - 6 \log 5 - 1) (10 \log 5 - 1) +

△ = 6 \log 3 - (10 \log 5 - 1) (6 \log 5 + 1) +

{(8 \log 5 - \log 3)}^{2}

z = \frac{(3 \log 3 + 1 - 2 \log 5) \pm \sqrt{6 \log 3 - (6 \log 5 + 1) (10 \log 5 - 1) + {(8 \log 5 - \log 3)}^{2}}}{2}

x = e^{(3 \log 3 - 2 \log 5 + 1) / 2} \times e^{\pm \sqrt{6 \log 3 + {(8 \log 5 - 3)}^{2} - 6 (\log 5 - 1) (6 kog 5 + 1)} / 2}

\Rightarrow Sum of roots is

r = e^{\frac{3}{2} \log 3 - \log 5 + \frac{1}{2}}

Commented by mr W last updated on 24/Oct/23

w r o n g!

x = e^{a \pm b}

\Rightarrow Σ x = e^{a} (e^{b} + e^{- b}) \neq e^{a}

Commented by a.lgnaoui last updated on 24/Oct/23

yes thanks

i use logaritme base 10

i think that my answer wil be exact

if the we change \log by \ln in the

given equation . (c^{'} est un piege!)

Commented by mr W last updated on 24/Oct/23

o u i!

Answered by mr W last updated on 24/Oct/23

\log_{3} x + \log_{3} (2, 5) - \log_{3} 3^{3} = \log_{x} 5 - \log_{x} 9

\log_{3} \frac{2.5 x}{27} = \log_{x} \frac{5}{9}

\frac{\ln x + \ln (\frac{2.5}{27})}{\log 3} = \frac{\ln \frac{5}{9}}{\ln x}

(\ln x) (\ln x + \ln \frac{2.5}{27}) = (\ln 3) (\ln \frac{5}{9})

l e t t = \ln x

t^{2} + (\ln \frac{2.5}{27}) t - (\ln 3) (\ln \frac{5}{9}) = 0

t_{1, 2} = \frac{1}{2} [- \ln \frac{2.5}{27} \pm \sqrt{{(\ln \frac{2.5}{27})}^{2} + 4 (\ln 3) (\ln \frac{5}{9})}]

t_{1} + t_{2} = - \ln \frac{2.5}{27} = \ln \frac{54}{5}

x_{1, 2} = e^{t_{1, 2}}

\underset{―}{s u m o f r o o t s :}

x_{1} + x_{2} = e^{t_{1}} + e^{t_{2}} = \dots .

\underset{―}{p r o d u c t o f r o o t s :}

x_{1} x_{2} = e^{t_{1}} e^{t_{2}} = e^{t_{1} + t_{2}} = e^{\ln \frac{54}{5}} = \frac{54}{5}

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