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

$$\mathrm{If} \\ $$$$\mathrm{x}^{\boldsymbol{\mathrm{log}}_{\mathrm{3}} \:\boldsymbol{\mathrm{x}}\:−\:\mathrm{1}} \:\:=\:\:\mathrm{log}_{\mathrm{3}} \:\mathrm{x}\:+\:\mathrm{3} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{log}_{\mathrm{3}} \:\mathrm{x}_{\mathrm{1}} \:+\:\mathrm{log}_{\mathrm{3}} \:\mathrm{x}_{\mathrm{2}} \:=\:? \\ $$

Commented by MATHEMATICSAM last updated on 16/Apr/24

$$\mathrm{Where}\:\mathrm{are}\:\mathrm{x}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{x}_{\mathrm{2}} \:\mathrm{in}\:\mathrm{your}\:\mathrm{equation}? \\ $$

Commented by hardmath last updated on 16/Apr/24

$$\mathrm{yes} \\ $$

Commented by Frix last updated on 16/Apr/24

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

x_1 and x_2 are the probably two solutions for the above euation

$$\mathrm{x}_{\mathrm{1}} \mathrm{and}\:\mathrm{x}_{\mathrm{2}} \:\mathrm{are}\:\mathrm{the}\:\mathrm{probably}\:\mathrm{two} \\ $$$$\mathrm{solutions}\:\mathrm{for}\:\mathrm{the}\:\mathrm{above}\:\mathrm{euation} \\ $$

Commented by MathematicalUser2357 last updated on 17/Apr/24

$${Where}'{s}\:“{q}''\:\mathrm{90}\:\leftarrow\:{for}\:{test} \\ $$

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

$$\mathrm{If}\:\mathrm{you}\:\mathrm{put}\:\mathrm{Q}/\mathrm{q}\:\mathrm{and}\:\mathrm{a}\:\mathrm{number}\:\mathrm{on}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{line}\:\mathrm{then}\:\mathrm{a}\:\mathrm{link}\:\mathrm{Q}\:\mathrm{number}\:\mathrm{is}\:\mathrm{automatically} \\ $$$$\mathrm{added}\:\mathrm{by}\:\mathrm{app} \\ $$

Commented by Frix last updated on 17/Apr/24

You can only approximate x_1 ≈.555575 x_2 ≈7.18458 log_3 x_1 +log_3 x_2 ≈1.25994

$$\mathrm{You}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$$${x}_{\mathrm{1}} \approx.\mathrm{555575} \\ $$$${x}_{\mathrm{2}} \approx\mathrm{7}.\mathrm{18458} \\ $$$$\mathrm{log}_{\mathrm{3}} \:{x}_{\mathrm{1}} \:+\mathrm{log}_{\mathrm{3}} \:{x}_{\mathrm{2}} \:\approx\mathrm{1}.\mathrm{25994} \\ $$

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)/ln3 witch gives: If x_1 and x_2 are the two solutions of: exp(ln(3)∙x ∙ (x−1)) = x+3 (eq_1 ) or x ∙ (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...

$$\mathrm{Okay},\:\mathrm{since}\:\mathrm{we}\:\mathrm{are}\:\mathrm{only}\:\mathrm{interested}\:\mathrm{in}\:\mathrm{the}\:\mathrm{logs} \\ $$$$\mathrm{of}\:\mathrm{x},\:\mathrm{we}\:\mathrm{can}\:\mathrm{reformulate}\:\mathrm{the}\:\mathrm{question}\:\mathrm{as}\:\mathrm{such}: \\ $$$${x}=\mathrm{log}_{\mathrm{3}} \mathrm{x}=\mathrm{ln}\left(\mathrm{x}\right)/\mathrm{ln3} \\ $$$$\mathrm{witch}\:\mathrm{gives}: \\ $$$$ \\ $$$$\mathrm{If}\:{x}_{\mathrm{1}} \:\mathrm{and}\:{x}_{\mathrm{2}} \:\mathrm{are}\:\mathrm{the}\:\mathrm{two}\:\mathrm{solutions}\:\mathrm{of}: \\ $$$$\mathrm{exp}\left(\mathrm{ln}\left(\mathrm{3}\right)\centerdot{x}\:\centerdot\:\left({x}−\mathrm{1}\right)\right)\:=\:{x}+\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{eq}_{\mathrm{1}} \right) \\ $$$$\mathrm{or} \\ $$$${x}\:\centerdot\:\left({x}−\mathrm{1}\right)\:=\:\mathrm{log}_{\mathrm{3}} \left({x}+\mathrm{3}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{eq}_{\mathrm{2}} \right) \\ $$$$\mathrm{Find}\:{x}_{\mathrm{1}} +{x}_{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{what}\:\mathrm{you}\:\mathrm{guys}\:\mathrm{think},\:\mathrm{but} \\ $$$$\mathrm{it}\:\mathrm{sound}\:\mathrm{a}\:\mathrm{bit}\:\mathrm{easier}\:\mathrm{to}\:\mathrm{deal}\:\mathrm{with}. \\ $$$$\mathrm{at}\:\mathrm{least}\:\mathrm{from}\:\mathrm{my}\:\mathrm{point}\:\mathrm{of}\:\mathrm{view}… \\ $$

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

x ∙ (x−1) = log_3 (x+3) can not be solved exactly, only approximately.

$${x}\:\centerdot\:\left({x}−\mathrm{1}\right)\:=\:\mathrm{log}_{\mathrm{3}} \left({x}+\mathrm{3}\right)\:{can}\:{not}\:{be} \\ $$$${solved}\:{exactly},\:{only}\:{approximately}. \\ $$

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

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