given that log3(2)=x, express log8(24) in terms of x

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∻Solution by Steps∻ ‖ step 1 ⋮ Given that $ \log_3(2) = x $, we need to express $ \log_8(24) $ in terms of $ x $. ‖ ‖ step 2 ⋮ We start by expressing $ \log_8(24) $ using the change of base formula: $ \log_8(24) = \frac{\log(24)}{\log(8)} $. ‖ ‖ step 3 ⋮ Next, we express $ \log(24) $ and $ \log(8) $ in terms of base 3: $ \log(24) = \log(3 \cdot 8) = \log(3) + \log(8) $ and $ \log(8) = \log(2^3) = 3 \log(2) $. ‖ ‖ step 4 ⋮ Substitute $ \log(8) $ and $ \log(24) $ into the change of base formula: $ \log_8(24) = \frac{\log(3) + \log(8)}{3 \log(2)} $. ‖ ‖ step 5 ⋮ Since $ \log(8) = 3 \log(2) $, we can rewrite the numerator: $ \log(3) + 3 \log(2) $. ‖ ‖ step 6 ⋮ Substitute $ \log(2) = x $ and $ \log(3) = \frac{\log(2)}{x} $: $ \log_8(24) = \frac{\frac{\log(2)}{x} + 3 \log(2)}{3 \log(2)} $. ‖ ‖ step 7 ⋮ Simplify the expression: $ \log_8(24) = \frac{\log(2) ( \frac{1}{x} + 3 )}{3 \log(2)} = \frac{1}{3} ( \frac{1}{x} + 3 ) $. ‖ ‖ step 8 ⋮ Further simplify: $ \log_8(24) = \frac{1}{3x} + 1 $. ‖ ∻Answer∻ ⚹ $ \log_8(24) = \frac{1}{3x} + 1 $ ⚹ ∻Key Concept∻ ⚹Change of Base Formula⚹ ∻Explanation∻ ⚹The change of base formula allows us to express logarithms in terms of a different base, which is useful for converting between different logarithmic expressions.⚹◊What is the relationship between log8(24) and log3(2)?⍭ Generate me a similar question◊

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