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Question Number 121387 by Ar Brandon last updated on 07/Nov/20

$$\frac{{\left(81\right)}^{1/{\log }_{5}9}+3^{3/{\log }_{\sqrt{6}}3}}{409}[{\left(\sqrt{7}\right)}^{2/{\log }_{25}7}-{\left(125\right)}^{{\log }_{25}6}]$$

Answered by 675480065 last updated on 07/Nov/20

$$\left(\frac{9^{{\log }_{9}25}+3^{{\log }_{3}6\sqrt{6}}}{409}\right)[7^{{\log }_{7}25}-5^{{\log }_{5}6\sqrt{6}}]$$$$=\left(\frac{25+6\sqrt{6}}{409}\right)(25-6\sqrt{6})$$$$=\left(\frac{1}{409}\right)({25}^2-36\left(6\right))=\frac{409}{409}=1$$

Answered by liberty last updated on 07/Nov/20

$$[{\left(9\right)}^{\log {}_9{\left(5\right)}^2}+3^{\log {}_3{\left(\sqrt{6}\right)}^3}][{\left(7\right)}^{\log {}_7\left(25\right)}-{\left(6\right)}^{\log {}_{25}\left(125\right)}]\times \frac{1}{409}$$$$=[25+6\sqrt{6}][25-6\sqrt{6}]\times \frac{1}{409}$$$$=(625-216)\times \frac{1}{409}=1$$$$625-216$$$$409.0$$

Commented by Ar Brandon last updated on 07/Nov/20

Cool. Thanks Sir

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