Question

Să se arate că 36 la puterea log in baza 6 a lui 5 +10 la puterea 1-lg2 - 3 la puterea log in baza 9 a lui 36=24

Answer 1

[tex]36^{log_65}+10^{1-lg2}-3^{log_936}= \\ \\ = (6^2)^{log_65}+10^{lg10-lg2}-3^{log_{3^2}36} \\ \\ = 6^{2log_65}+10^{lg \frac{10}{2} }-3^{ \frac{1}{2} log_{3}36} \\ \\ = 6^{log_65^2}+10^{lg 5 }-3^{ log_{3}36^\frac{1}{2} } \\ \\ = 6^{log_625}+5-3^{ log_{3} \sqrt{36} } \\ \\ = 25+5-3^{ log_{3} 6 } \\ \\ = 25+5-6 \\ \\ = 24[/tex]