Discussion on Exponentials and Logarithms


 

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Posted on: 09:03:06 AM on July 6, 2015
Posted By:
Nikita Masand
How should I solve, (2^log base 6 18) (3^log base 6 3) Ans is 6


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Posted on: 12:17:52 AM on October 26, 2015
(2^log base 6 18) (3^log base 6 3)



Posted on: 12:25:44 AM on October 26, 2015
(2^log base 6 18) (3^log base 6 3) =((2^log base 6 (6)*(3)) (3^log base 6 3) =(2^((log base 6 6)+ (log base 6 3))(3^log base 6 3) [using log base x (ab) = log base x a + log base x b] =(2^(1 + (log base 6 3))(3^log base 6 3) =(2)*(2)^(log base 6 3)(3^log base 6 3)...................(1) Now, we know that f(x)^(log base a g(x))= g(x)^(log base a f(x)) Therefore, equation (1) can be written as, =(2)*(3)^(log base 6 2)(3^log base 6 3) =(2)*(3)^((log base 6 2) + (log base 6 3)) ....... [using (a^m)*(a^n)=a*(m+n)] =(2)*(3)^((log base 6 (2)*(3)) [using log base x a + log base x b = log base x (ab)] =(2)*(3)^((log base 6 6) =(2)*(3)^(1) =6 (Ans.)



Posted on: 11:40:47 PM on April 10, 2017
{2^(log base 6 18)} * {3^(log base 6 3)} {18^(log base 6 2)} * {3^(log base 6 3)} -----since{a^(log base b c)}=={c^(log base b a)} {(3*6)^(log base 6 2)} * {3^(log base 6 3)} {(3^log base 6 2)*(6^log base 6 2)} * {3^(log base 6 3)} {(3^log base 6 2)* 2} * {3^(log base 6 3)} -------since {a^log base a x==x} 2*{(3^log base 6 2) * 3^(log base 6 3)} 2*{3^(log base 6 2 + log base 6 3)} 2*{3^(log base 6 (2*3)} 2*{3^(log base 6 6)} 2*{3^1} 2*3 6





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