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Question Number 17386 by ajfour last updated on 05/Jul/17

$$\mathrm{Solve}\mathrm{the}\mathrm{equation}:$$$$\log {}_2\mathrm{x}\log {}_3\mathrm{x}\log {}_5\mathrm{x}=\log {}_2\mathrm{x}\log {}_3\mathrm{x}$$$$+\log {}_3\mathrm{x}\log {}_5\mathrm{x}$$$$+\log {}_5\mathrm{x}\log {}_2\mathrm{x}.$$

Answered by mrW1 last updated on 05/Jul/17

$$\mathrm{let}\mathrm{u}=\ln \mathrm{x}$$$$\mathrm{a}=\ln 2$$$$\mathrm{b}=\ln 3$$$$\mathrm{c}=\ln 5$$$$\Rightarrow \frac{{\mathrm{u}}^3}{\mathrm{abc}}=\frac{{\mathrm{u}}^2}{\mathrm{ab}}+\frac{{\mathrm{u}}^2}{\mathrm{bc}}+\frac{{\mathrm{u}}^2}{\mathrm{ca}}$$$$\Rightarrow {\mathrm{u}}^3=(\mathrm{a}+\mathrm{b}+\mathrm{c}){\mathrm{u}}^2$$$$\Rightarrow {\mathrm{u}}^2[\mathrm{u}-(\mathrm{a}+\mathrm{b}+\mathrm{c})]=0$$$$\Rightarrow \mathrm{u}=0$$$$\mathrm{or}$$$$\Rightarrow \mathrm{u}=\mathrm{a}+\mathrm{b}+\mathrm{c}$$$$$$$$\Rightarrow \ln \mathrm{x}=0$$$$\Rightarrow \mathrm{x}=1$$$$$$$$\Rightarrow \ln \mathrm{x}=\ln 2+\ln 3+\ln 5=\ln \left(2\times 3\times 5\right)$$$$\Rightarrow \mathrm{x}=2\times 3\times 5=30$$

Commented by ajfour last updated on 05/Jul/17

$$\mathrm{great},\mathrm{you}\mathrm{play}\mathrm{confident}\mathrm{sir}!$$$$\mathrm{who}\mathrm{else}\mathrm{can}?$$

Commented by 47358857 last updated on 07/Dec/17

Answered by b.e.h.i.8.3.4.17@gmail.com last updated on 06/Jul/17

$$abc=ab+bc+ca$$$$\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\Rightarrow {log}_x^2+{log}_x^3+{log}_x^5=1$$$$\Rightarrow {log}_x^{2\times 3\times 5}=1\Rightarrow {log}_x^{30}=1\Rightarrow x=30.$$$$also:x=1,canbeananswer.$$

Commented by ajfour last updated on 08/Jul/17

$$\mathrm{very}\mathrm{nice}\mathrm{way},\mathrm{sir}.$$

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