If-x-log-3-x-1-log-3-x-3-Find-log-3-x-1-log-3-x-2-

Question Number 206501 by hardmath last updated on 16/Apr/24

If

x^{\log_{3} x - 1} = \log_{3} x + 3

Find : \log_{3} x_{1} + \log_{3} x_{2} = ?

Commented by MATHEMATICSAM last updated on 16/Apr/24

Where are x_{1} and x_{2} in your equation ?

Commented by hardmath last updated on 16/Apr/24

yes

Commented by Frix last updated on 16/Apr/24

��

Commented by Tinku Tara last updated on 17/Apr/24

x_{1} and x_{2} are the probably two

solutions for the above euation

Commented by MathematicalUser2357 last updated on 17/Apr/24

{W h e r e}^{'} s “ q^{″} 90 \leftarrow f o r t e s t

Commented by Tinku Tara last updated on 17/Apr/24

If you put Q / q and a number on the same

line then a link Q number is automatically

added by app

Commented by Frix last updated on 17/Apr/24

You can only approximate

x_{1} \approx . 555575

x_{2} \approx 7.18458

\log_{3} x_{1} + \log_{3} x_{2} \approx 1.25994

Commented by TheHoneyCat last updated on 20/Apr/24

Okay, since we are only interested in the logs

of x, we can reformulate the question as such :

x = \log_{3} x = \ln (x) / \ln 3

witch gives :

If x_{1} and x_{2} are the two solutions of :

\exp (\ln (3) \cdot x \cdot (x - 1)) = x + 3 ({eq}_{1})

or

x \cdot (x - 1) = \log_{3} (x + 3) ({eq}_{2})

Find x_{1} + x_{2}

I {don}^{'} t know what you guys think, but

it sound a bit easier to deal with .

at least from my point of view \dots

Commented by mr W last updated on 20/Apr/24

x \cdot (x - 1) = \log_{3} (x + 3) c a n n o t b e

s o l v e d e x a c t l y, o n l y a p p r o x i m a t e l y .

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