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a) log4 (x+8) = log2 3 b) 3 ^2x= 5/3x c) log3 x = 2-log3 (x+8) d) log3 (2y-4) = 2 log3 2
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\[a) \log_{4} (x+8) = \log_{2} 3\]
\[b) 3^{2x} = 5 / 3^{x}\]
c) \[\log_{3} x=2-\log_{3} (x+8)\]
\[\log_{3} (2y-4) = 2 \log_{3} 2 \]
= log3 x2 + log3 5 = log3(5x2)
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@Callisto @FoolForMath @apoorvk @aroub
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