please help me with this question:prove that 7log16\15+5log25\24+3 log81\80=log2
what is the 9th term if the nth term is n(n+1)\3 )
what is the arithmetic mean of 34 and 44?
Question #1 hint: \[\begin{gathered} \log \left( {\frac{{16}}{{25}}} \right) = \log 16 - \log 25 = \log \left( {{2^4}} \right) - \log \left( {{5^2}} \right) = 4\log 2 - 2\log 5 \hfill \\ \log \left( {\frac{{25}}{{24}}} \right) = \log 25 - \log 24 = \log \left( {{5^2}} \right) - \log \left( {{2^3} \cdot 3} \right) = 2\log 5 - 3\log 2 - \log 3 \hfill \\ \log \left( {\frac{{81}}{{80}}} \right) = \log 81 - \log 80 = \log \left( {{3^4}} \right) - \log \left( {{2^4} \cdot 5} \right) = 4\log 3 - 4\log 2 - \log 5 \hfill \\ \end{gathered} \]
question #3 what is: \[\frac{{34 + 44}}{2} = ...?\]
oho its not 16/25 its 16/15
oops.. you are right! \[\begin{gathered} \log \left( {\frac{{16}}{{15}}} \right) = \log 16 - \log 15 = \log \left( {{2^4}} \right) - \log \left( {3 \cdot 5} \right) = 4\log 2 - \log 3 - \log 5 \hfill \\ \log \left( {\frac{{25}}{{24}}} \right) = \log 25 - \log 24 = \log \left( {{5^2}} \right) - \log \left( {{2^3} \cdot 3} \right) = 2\log 5 - 3\log 2 - \log 3 \hfill \\ \log \left( {\frac{{81}}{{80}}} \right) = \log 81 - \log 80 = \log \left( {{3^4}} \right) - \log \left( {{2^4} \cdot 5} \right) = 4\log 3 - 4\log 2 - \log 5 \hfill \\ \end{gathered} \]
thank you so much
thank you!
I've another doubt please clarify !what is A.G.P(arithmetic geometric progression).
keep in mind that there are two progressions, namely 1) the arithmetic progression 2) the geometric progression
we have an arithmetic progression when the difference between one term and its consecutive term (previous or subsequent) is constant
whereas we hae a geometri progression, where the quotient between one term and its consective term (previous or subsequent) is costant
thank you for explaining me about progression now I can manage progressions
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