How do you condense 1/2(log_4(x+1)+2log_4(x-1))+6log_4x12(log4(x+1)+2log4(x−1))+6log4x?

1 Answer
Carlos S.
Jan 18, 2018

Let E = 1/2[log_4(x+1) + 2log_4(x-1)] + 6log_4xE=12[log4(x+1)+2log4(x−1)]+6log4x represent your expression (just so then I don't have to rewrite it so often!). Using the logarithm's identities:

E = 1/2[log_4(x+1)+ log_4(x-1)^2] + log_4x^6E=12[log4(x+1)+log4(x−1)2]+log4x6;

E = 1/2log_4[(x+1)(x-1)^2] + log_4x^6E=12log4[(x+1)(x−1)2]+log4x6;

E = log_4[(x+1)^(1/2)(x-1)] + log_4x^6E=log4[(x+1)12(x−1)]+log4x6;

E = log_4[(x+1)^(1/2)(x-1)x^6]E=log4[(x+1)12(x−1)x6].

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