log-24-3-a-and-log-24-6-b-6-log-8-b-4a-

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

\log_{24} 3 = a and \log_{24} 6 = \frac{b}{6}

\log_{\sqrt{8}} (b - 4 a) = ?

Commented by oubiji last updated on 19/Jan/25

b - 4 a = \log_{24} 6^{6} - {l o g}_{24} 3^{4}

= {l o g}_{24} \frac{6^{6}}{3^{4}}

= {l o g}_{24} 24^{2}

= 2

t h e n :

{l o g}_{\sqrt{8}} (b - 4 a) = {l o g}_{\sqrt{8}} 2

= \frac{l n 2}{l n \sqrt{8}}

= \frac{l n 2}{\frac{3}{2} l n 2}

= \frac{2}{3}

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

p l e a s e n e x t t i m e p o s t y o u r s o l u t i o n a s

“ {a n s w e r}^{″} n o t a s a “ {c o m m e n t}^{″}

Answered by A5T last updated on 19/Jan/25

b = \log_{24} 6^{6}; 4 a = \log_{24} 3^{4}

b - 4 a = \log_{24} (\frac{6^{6}}{3^{4}}) = \log_{24} (24^{2}) = 2

\Rightarrow \log_{\sqrt{8}} (b - 4 a) \Rightarrow \log_{\sqrt{8}} (2) = \log_{2^{\frac{3}{2}}} (2) = \frac{2}{3}

Answered by MathematicalUser2357 last updated on 26/Jan/25

\log_{\sqrt{8}} (6 \times \log_{24} 6 - \log_{24} 3)

FAILED TO CALCULATE

why do tinku tara {can}^{'} t do \log_{b} x