To evaluate logarithmic expressions, we use the property that logb(ac)=c⋅logb(a)\log_b(a^c) = c \cdot \log_b(a)logb(ac)=c⋅logb(a). For expressions like logb(bc)\log_b(b^c)logb(bc), the result is simply ccc because logb(b)=1\log_b(b) = 1logb(b)=1. For logb(a)\log_b(a)logb(a) where aaa is a power of bbb, we find the exponent that makes bbb equal to aaa.
Using the property of logarithms, we have: log2(211)=11 \log_{2}(2^{11}) = 11 log2(211)=11
Recognizing that 243=35 243 = 3^{5} 243=35, we can express the logarithm as: log3(243)=log3(35)=5 \log_{3}(243) = \log_{3}(3^{5}) = 5 log3(243)=log3(35)=5 However, due to numerical precision, the computed value is approximately 4.999999999999999 4.999999999999999 4.999999999999999, which we round to 5 5 5.
Since 256=44 256 = 4^{4} 256=44, we find: log4(256)=log4(44)=4 \log_{4}(256) = \log_{4}(4^{4}) = 4 log4(256)=log4(44)=4
The results for each expression are:
Thus, the final answers are: 11,5,4 \boxed{11}, \quad \boxed{5}, \quad \boxed{4} 11,5,4
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