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Question Number 222679 by fantastic last updated on 04/Jul/25
If x=Π_(x=1) ^(10) x then (1/(log _2 x))+(1/(log _3 x))+(1/(log _4 x))...+(1/(log _(10) x))=??
$${If}\:{x}=\underset{{x}=\mathrm{1}} {\overset{\mathrm{10}} {\prod}}{x}\:{then}\:\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{2}} {x}}+\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{3}} {x}}+\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{4}} {x}}…+\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{10}} {x}}=?? \\ $$
Answered by Tawa11 last updated on 04/Jul/25
x = 1 × 2 × 3 × .. × 10 = 10! log_x 1 + log_x 2 + log_x 3 + ... + log_x 10 log_x (1 × 2 × 3 × ... × 10) = log_(10!) 10! = 1
$$\mathrm{x}\:=\:\mathrm{1}\:×\:\mathrm{2}\:×\:\mathrm{3}\:×\:..\:×\:\mathrm{10}\:\:\:=\:\:\:\mathrm{10}! \\ $$$$\mathrm{log}_{\mathrm{x}} \mathrm{1}\:\:+\:\:\mathrm{log}_{\mathrm{x}} \mathrm{2}\:\:+\:\:\mathrm{log}_{\mathrm{x}} \mathrm{3}\:\:+\:\:…\:\:+\:\:\mathrm{log}_{\mathrm{x}} \mathrm{10} \\ $$$$\mathrm{log}_{\mathrm{x}} \left(\mathrm{1}\:×\:\mathrm{2}\:×\:\mathrm{3}\:×\:…\:×\:\mathrm{10}\right)\:\:\:=\:\:\:\mathrm{log}_{\mathrm{10}!} \mathrm{10}!\:\:\:=\:\:\:\mathrm{1} \\ $$
Answered by MrGaster last updated on 05/Jul/25
(1)x=Π_(k=1) ^(10) k=10! Σ_(k=2) ^(10) (1/(log_k x))=Σ_(k=2) ^(10) log_x k Σ_(k=2) ^(10) log_x k=log_x (Π_(k=2) ^(10) k) Π_(k=2) ^(10) k=((10)/1)=10! log_x (10)! x=10⇒log_(10!) (10!)=1 (2)x=Π_(k=1) ^(10) k Σ_(m=2) ^(10) (1/(log_m x))=Σ_(m=2) ^(10) log_x m Σ_(m=2) ^(10) log_m m=log_x (Π_(m=2) ^(10) m) Π_(m=2) ^(10) m=((Π_(k=1) ^(10) k)/1) ((Π_(k=1) ^(10) k)/1)=(x/1) log_x (Π_(m=2) ^(10) m)=log_x ((x/1)) log((x/1))=log_x x−log_x 1 log_x x=1 log_x 1==0 1−0=1
$$\left(\mathrm{1}\right){x}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{10}} {\prod}}{k}=\mathrm{10}! \\ $$$$\underset{{k}=\mathrm{2}} {\overset{\mathrm{10}} {\sum}}\frac{\mathrm{1}}{\mathrm{log}_{{k}} {x}}=\underset{{k}=\mathrm{2}} {\overset{\mathrm{10}} {\sum}}\mathrm{log}_{{x}} {k} \\ $$$$\underset{{k}=\mathrm{2}} {\overset{\mathrm{10}} {\sum}}\mathrm{log}_{{x}} {k}=\mathrm{log}_{{x}} \left(\underset{{k}=\mathrm{2}} {\overset{\mathrm{10}} {\prod}}{k}\right) \\ $$$$\underset{{k}=\mathrm{2}} {\overset{\mathrm{10}} {\prod}}{k}=\frac{\mathrm{10}}{\mathrm{1}}=\mathrm{10}! \\ $$$$\mathrm{log}_{{x}} \left(\mathrm{10}\right)! \\ $$$${x}=\mathrm{10}\Rightarrow\mathrm{log}_{\mathrm{10}!} \left(\mathrm{10}!\right)=\mathrm{1} \\ $$$$\left(\mathrm{2}\right){x}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{10}} {\prod}}{k} \\ $$$$\underset{{m}=\mathrm{2}} {\overset{\mathrm{10}} {\sum}}\frac{\mathrm{1}}{\mathrm{log}_{{m}} {x}}=\underset{{m}=\mathrm{2}} {\overset{\mathrm{10}} {\sum}}\mathrm{log}_{{x}} {m} \\ $$$$\underset{{m}=\mathrm{2}} {\overset{\mathrm{10}} {\sum}}\mathrm{log}_{{m}} {m}=\mathrm{log}_{{x}} \left(\underset{{m}=\mathrm{2}} {\overset{\mathrm{10}} {\prod}}{m}\right) \\ $$$$\underset{{m}=\mathrm{2}} {\overset{\mathrm{10}} {\prod}}{m}=\frac{\underset{{k}=\mathrm{1}} {\overset{\mathrm{10}} {\prod}}{k}}{\mathrm{1}} \\ $$$$\frac{\underset{{k}=\mathrm{1}} {\overset{\mathrm{10}} {\prod}}{k}}{\mathrm{1}}=\frac{{x}}{\mathrm{1}} \\ $$$$\mathrm{log}_{{x}} \left(\underset{{m}=\mathrm{2}} {\overset{\mathrm{10}} {\prod}}{m}\right)=\mathrm{log}_{{x}} \left(\frac{{x}}{\mathrm{1}}\right) \\ $$$$\mathrm{log}\left(\frac{{x}}{\mathrm{1}}\right)=\mathrm{log}_{{x}} {x}−\mathrm{log}_{{x}} \mathrm{1} \\ $$$$\mathrm{log}_{{x}} {x}=\mathrm{1} \\ $$$$\mathrm{log}_{{x}} \mathrm{1}==\mathrm{0} \\ $$$$\mathrm{1}−\mathrm{0}=\mathrm{1} \\ $$

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