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1-2-log-3-25-1-2-log-5-9-




Question Number 146895 by mathdanisur last updated on 16/Jul/21
(1/(2 + log_3 (25))) + (1/(2 + log_5 (9))) = ?
$$\frac{\mathrm{1}}{\mathrm{2}\:+\:\boldsymbol{{log}}_{\mathrm{3}} \left(\mathrm{25}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{2}\:+\:\boldsymbol{{log}}_{\mathrm{5}} \left(\mathrm{9}\right)}\:=\:? \\ $$
Answered by EDWIN88 last updated on 16/Jul/21
⇒ L =(1/(2+log _3 (25)))+(1/(2+log _5 (9)))  ⇒L=(1/(log _3 (9×25))) +(1/(log _5 (9×25)))  ⇒L=log _(225) (3)+log _(225) (5)  ⇒L=log _(225) (15) = (1/2)
$$\Rightarrow\:\mathrm{L}\:=\frac{\mathrm{1}}{\mathrm{2}+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{25}\right)}+\frac{\mathrm{1}}{\mathrm{2}+\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{9}\right)} \\ $$$$\Rightarrow\mathrm{L}=\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{9}×\mathrm{25}\right)}\:+\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{9}×\mathrm{25}\right)} \\ $$$$\Rightarrow\mathrm{L}=\mathrm{log}\:_{\mathrm{225}} \left(\mathrm{3}\right)+\mathrm{log}\:_{\mathrm{225}} \left(\mathrm{5}\right) \\ $$$$\Rightarrow\mathrm{L}=\mathrm{log}\:_{\mathrm{225}} \left(\mathrm{15}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mathdanisur last updated on 16/Jul/21
thank you Ser
$${thank}\:{you}\:{Ser} \\ $$

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