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Question Number 183777 by Michaelfaraday last updated on 30/Dec/22

Commented by Frix last updated on 30/Dec/22

$$k=2\sqrt[3]{3}$$

Answered by HeferH last updated on 30/Dec/22

$${24}^{1-\frac{1}{k}}=3$$$$\frac{24}{{24}^{\frac{1}{k}}}=3$$$$24=3\cdot {24}^{\frac{1}{k}}$$$$8={24}^{\frac{1}{k}}$$$$2^{3k}=24$$$$2^k=2\sqrt[3]{3}$$

Commented by Michaelfaraday last updated on 30/Dec/22

$$thankssir$$

Answered by cortano1 last updated on 30/Dec/22

$$1-\frac{1}{k}=\log {}_{24}\left(3\right)$$$$\Rightarrow \frac{1}{k}=1-\log {}_{24}\left(3\right)=\log {}_{24}\left(8\right)$$$$\Rightarrow k=\log {}_8\left(24\right)=\frac{1}{3}\{3+\log {}_{2}3\}$$$$\therefore 2^k=2^{1+\log {}_2(\left(\sqrt[3]{3}\right)}=2.\sqrt[3]{3}$$$$$$$$$$

Commented by Michaelfaraday last updated on 30/Dec/22

$$thankssir$$

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