Question

7log base 10 (15÷16)+6 log base 10 (8÷3) + 5 log base 10 (2÷5) + log base 10 (32÷25)=?​

Answer 1

Required Answer:-

Given to Evaluate:

$\rm \mapsto 7 log_{10} \bigg( \dfrac{15}{16} \bigg) + 6 log_{10} \bigg( \dfrac{8}{3} \bigg) + 5 log_{10} \bigg( \dfrac{2}{5} \bigg) + log_{10} \bigg( \dfrac{32}{25} \bigg)$

Solution:

Given,

$\rm 7 log_{10} \bigg( \dfrac{15}{16} \bigg) + 6 log_{10} \bigg( \dfrac{8}{3} \bigg) + 5 log_{10} \bigg( \dfrac{2}{5} \bigg) + log_{10} \bigg( \dfrac{32}{25} \bigg)$

$\rm = log_{10} \bigg( \dfrac{15}{16} \bigg)^{7} +log_{10} \bigg( \dfrac{8}{3} \bigg)^{6} + log_{10} \bigg( \dfrac{2}{5} \bigg) ^{5} + log_{10} \bigg( \dfrac{32}{25} \bigg)$

$\rm = log_{10} \bigg( \dfrac{ {15}^{7} }{ {16}^{7} } \times \dfrac{ {8}^{6} }{ {3}^{6} } \times \dfrac{ {2}^{5} }{ {5}^{5} } \times \dfrac{32}{25} \bigg)$

$\rm = log_{10} \bigg( \dfrac{ {(3 \times 5)}^{7} }{ {2}^{4 \times 7} } \times \dfrac{ {2}^{3 \times 6} }{ {3}^{6} } \times \dfrac{ {2}^{5} }{ {5}^{5} } \times \dfrac{ {2}^{5} }{ {5}^{2} } \bigg)$

$\rm = log_{10} \bigg( \dfrac{ {3 ^{7} \times 5}^{7} }{ {2}^{28} } \times \dfrac{ {2}^{18} }{ {3}^{6} } \times \dfrac{ {2}^{5} }{ {5}^{5} } \times \dfrac{ {2}^{5} }{ {5}^{2} } \bigg)$

$\rm = log_{10} \bigg( \dfrac{ {3 ^{7} \times \cancel{5}^{7} }}{ {2}^{28} \times \cancel{{5}^{7} } } \times \dfrac{ {2}^{18 + 5 + 5} }{ {3}^{6} } \bigg)$

$\rm = log_{10} \bigg( \dfrac{ {3 ^{7} }}{ \cancel{ {2}^{28} }} \times \dfrac{ \cancel{{2}^{28}} }{ {3}^{6} } \bigg)$

$\rm = log_{10} \bigg( \dfrac{ {3 ^{7} }}{ {3}^{6} }\bigg)$

$\rm = log_{10} (3)$

Hence, the simplified form is log(3)

Answer:

• The simplified form is log(3)

Formula Used:

$\rm \mapsto n \log_{x}(y) = log_{x}(y^{n})$

$\rm \mapsto log(x) + log(y) = log(xy)$

Answer 2

Answer:

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