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0-16-log-2-5-1-3-1-3-2-1-3-3-




Question Number 116491 by bemath last updated on 04/Oct/20
(0.16)^(log _(2.5) ((1/3)+(1/3^2 )+(1/3^3 )+...))  =?
$$\left(\mathrm{0}.\mathrm{16}\right)^{\mathrm{log}\:_{\mathrm{2}.\mathrm{5}} \left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+…\right)} \:=? \\ $$
Answered by bobhans last updated on 04/Oct/20
Only applying property of logarithm  ⇒ a^(log _a  f(x))  = f(x)  ⇒((2/5))^(2.log _(2.5) ((1/3)+(1/3^2 )+(1/3^3 )+...)) = ((1/3)+(1/3^2 )+(1/3^3 )+...)^(−2)       = (((1/3)/(1−(1/3))))^(−2) = ((1/2))^(−2)  = (2)^2  = 4
$$\mathrm{Only}\:\mathrm{applying}\:\mathrm{property}\:\mathrm{of}\:\mathrm{logarithm} \\ $$$$\Rightarrow\:\mathrm{a}^{\mathrm{log}\:_{\mathrm{a}} \:\mathrm{f}\left(\mathrm{x}\right)} \:=\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\Rightarrow\left(\frac{\mathrm{2}}{\mathrm{5}}\right)^{\mathrm{2}.\mathrm{log}\:_{\mathrm{2}.\mathrm{5}} \left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+…\right)} =\:\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+…\right)^{−\mathrm{2}} \\ $$$$\:\:\:\:=\:\left(\frac{\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}\right)^{−\mathrm{2}} =\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\mathrm{2}} \:=\:\left(\mathrm{2}\right)^{\mathrm{2}} \:=\:\mathrm{4} \\ $$

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