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Question Number 215840 by golsendro last updated on 19/Jan/25

$$\log {}_{24}3=a\mathrm{and}\log {}_{24}6=\frac{b}{6}$$$$\log {}_{\sqrt{8}}(b-4a)=?$$

Commented by oubiji last updated on 19/Jan/25

$$b-4a={\log }_{24}{6}^6-{log}_{24}{3}^4$$$$={log}_{24}\frac{6^6}{3^4}$$$$={log}_{24}{24}^2$$$$=2$$$$then:$$$${log}_{\sqrt{8}}(b-4a)={log}_{\sqrt{8}}2$$$$=\frac{ln2}{ln\sqrt{8}}$$$$=\frac{ln2}{\frac{3}{2}ln2}$$$$=\frac{2}{3}$$$$$$$$$$$$$$$$$$

Commented by Rasheed.Sindhi last updated on 20/Jan/25

$$pleasenexttimepostyoursolutionas$$$$‘‘{answer}^{″}notasa‘‘{comment}^{″}$$

Answered by A5T last updated on 19/Jan/25

$$\mathrm{b}={\log }_{24}{6}^6;4\mathrm{a}={\log }_{24}{3}^4$$$$\mathrm{b}-4\mathrm{a}={\log }_{24}\left(\frac{6^6}{3^4}\right)={\log }_{24}\left({24}^2\right)=2$$$$\Rightarrow {\log }_{\sqrt{8}}(\mathrm{b}-4\mathrm{a})\Rightarrow {\log }_{\sqrt{8}}\left(2\right)={\log }_{2^{\frac{3}{2}}}\left(2\right)=\frac{2}{3}$$

Answered by MathematicalUser2357 last updated on 26/Jan/25

$${\log }_{\sqrt{8}}(6\times {\log }_{24}6-{\log }_{24}3)$$$$\mathrm{FAILED}\mathrm{TO}\mathrm{CALCULATE}$$$$\mathrm{why}\mathrm{do}\mathrm{tinku}\mathrm{tara}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{do}{\log }_{b}x$$

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