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

$$\mathrm{If}$$$${\mathrm{x}}^{{\mathbf{log}}_3\mathbf{x}-1}={\log }_3\mathrm{x}+3$$$$\mathrm{Find}:{\log }_3{\mathrm{x}}_1+{\log }_3{\mathrm{x}}_2=?$$

Commented by MATHEMATICSAM last updated on 16/Apr/24

$$\mathrm{Where}\mathrm{are}{\mathrm{x}}_1\mathrm{and}{\mathrm{x}}_2\mathrm{in}\mathrm{your}\mathrm{equation}?$$

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

$${\mathrm{x}}_1\mathrm{and}{\mathrm{x}}_2\mathrm{are}\mathrm{the}\mathrm{probably}\mathrm{two}$$$$\mathrm{solutions}\mathrm{for}\mathrm{the}\mathrm{above}\mathrm{euation}$$

Commented by MathematicalUser2357 last updated on 17/Apr/24

$${Where}^{\prime }s‘‘q^{″}90\leftarrow fortest$$

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

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

Commented by Frix last updated on 17/Apr/24

$$\mathrm{You}\mathrm{can}\mathrm{only}\mathrm{approximate}$$$$x_1\approx .555575$$$$x_2\approx 7.18458$$$${\log }_3{x}_1+{\log }_3{x}_2\approx 1.25994$$

Commented by TheHoneyCat last updated on 20/Apr/24

$$\mathrm{Okay},\mathrm{since}\mathrm{we}\mathrm{are}\mathrm{only}\mathrm{interested}\mathrm{in}\mathrm{the}\mathrm{logs}$$$$\mathrm{of}\mathrm{x},\mathrm{we}\mathrm{can}\mathrm{reformulate}\mathrm{the}\mathrm{question}\mathrm{as}\mathrm{such}:$$$$x={\log }_3\mathrm{x}=\ln \left(\mathrm{x}\right)/\ln 3$$$$\mathrm{witch}\mathrm{gives}:$$$$$$$$\mathrm{If}{x}_1\mathrm{and}{x}_2\mathrm{are}\mathrm{the}\mathrm{two}\mathrm{solutions}\mathrm{of}:$$$$\exp (\ln \left(3\right)\cdot x\cdot (x-1))=x+3\left({\mathrm{eq}}_1\right)$$$$\mathrm{or}$$$$x\cdot (x-1)={\log }_3(x+3)\left({\mathrm{eq}}_2\right)$$$$\mathrm{Find}{x}_1+x_2$$$$$$$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{what}\mathrm{you}\mathrm{guys}\mathrm{think},\mathrm{but}$$$$\mathrm{it}\mathrm{sound}\mathrm{a}\mathrm{bit}\mathrm{easier}\mathrm{to}\mathrm{deal}\mathrm{with}.$$$$\mathrm{at}\mathrm{least}\mathrm{from}\mathrm{my}\mathrm{point}\mathrm{of}\mathrm{view}...$$

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

$$x\cdot (x-1)={\log }_3(x+3)cannotbe$$$$solvedexactly,onlyapproximately.$$

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