Question

log base 10 2+ 16 log base 10 (16÷15) + 12 log base 10 (25÷24) + 7 log base 10 (81÷80)=?​

Answer 1

Answer:

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Answer 2

Required Answer:-

Given to evaluate:

$\rm \mapsto log_{10}(2) + 16 log_{10} \bigg( \dfrac{16}{15} \bigg) + 12 log_{10} \bigg( \dfrac{25}{24} \bigg) + 7 log_{10} \bigg( \dfrac{81}{80} \bigg)$

Solution:

$\rm log_{10}(2) + 16 log_{10} \bigg( \dfrac{16}{15} \bigg) + 12 log_{10} \bigg( \dfrac{25}{24} \bigg) + 7 log_{10} \bigg( \dfrac{81}{80} \bigg)$

$\rm = log_{10}(2) + log_{10} \bigg( \dfrac{ {16}^{16} }{ {15}^{16} } \bigg) + log_{10} \bigg( \dfrac{ {25}^{12} }{ {24}^{12} } \bigg) + log_{10} \bigg( \dfrac{ {81}^{7} }{ {80}^{7} } \bigg)$

$\rm = log_{10}(2) + log_{10} \bigg( \dfrac{ {2}^{4 \times 16} }{ {(3 \times 5)}^{16} } \bigg) + log_{10} \bigg( \dfrac{ {5}^{12 \times 2} }{ {(3 \times {2}^{3}) }^{12} } \bigg) + log_{10} \bigg( \dfrac{ {3}^{4 \times 7} }{ {( {2}^{4} \times 5)}^{7} } \bigg)$

$\rm = log_{10}(2) + log_{10} \bigg( \dfrac{ {2}^{64} }{ {3}^{16} \times 5^{16} } \bigg) + log_{10} \bigg( \dfrac{ {5}^{24} }{ {3}^{12} \times {2}^{36}} \bigg) + log_{10} \bigg( \dfrac{ {3}^{28} }{{2}^{28} \times 5^{7}} \bigg)$

$\rm = log_{10} \bigg( \dfrac{ {2}^{64} \times 2 }{ {3}^{16} \times 5^{16} } \times \dfrac{ {5}^{24} }{ {3}^{12} \times {2}^{36}} \times \dfrac{ {3}^{28} }{{2}^{28} \times 5^{7}} \bigg)$

$\rm = log_{10} \bigg( \dfrac{ {2}^{65} }{ {3}^{28} \times 5^{23} } \times \dfrac{ {5}^{24} }{ {2}^{64}} \times {3}^{28} \bigg)$

$\rm = log_{10} \bigg( {2}^{65 - 64} \times \ {5}^{24 - 23}\bigg)$

$\rm = log_{10} ( 2\times 5)$

$\rm = log_{10}(10 )$

$\rm = 1$

Hence, the required answer is 1