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Question Number 136054 by azadsir last updated on 18/Mar/21

Commented by mr W last updated on 18/Mar/21

a=log_x  xyz=(1/(log_(xyz)  x))  ⇒(1/a)=log_(xyz)  x  ⇒(1/a)+(1/b)+(1/c)=log_(xyz)  x+log_(xyz)  y+log_(xyz)  z                     =log_(xyz)  xyz=1

$${a}=\mathrm{log}_{{x}} \:{xyz}=\frac{\mathrm{1}}{\mathrm{log}_{{xyz}} \:{x}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}}=\mathrm{log}_{{xyz}} \:{x} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\mathrm{log}_{{xyz}} \:{x}+\mathrm{log}_{{xyz}} \:{y}+\mathrm{log}_{{xyz}} \:{z} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{log}_{{xyz}} \:{xyz}=\mathrm{1} \\ $$

Commented by otchereabdullai@gmail.com last updated on 19/Mar/21

wow!

$$\mathrm{wow}! \\ $$

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