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\log _{2}^{x}+\log _{4}^{x}+\log _{16}^{x}=\frac{21}{4} |
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Answer» log base 2 x = log(x) / log(2)log base 4 x = log(x) / log(4) = log(x) / log(2^2) = log(x) / ( 2 log(2) )log base 16 x = log(x) / log(16) = log(x) / log(2^4) = log(x) / (4 log(2) ) log base 2 x + log base 4 x + log base 16 x = log(x) / log(2) + log(x) / ( 2 log(2)) + log(x) / ( 4 log(2))= (log(x) / log(2) )( 1 + 1/2+1/4)= ( 1 + 1/2+1/4) (log(x) / log(2) )= (7/4) log (base 2) x = 21/4multiply both sides by 47 log(base 2) x = 21divide both sides by 7log(base 2) x = 3x = 2^3x = 8 |
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