How do you condense #1/2(log_4(x+1)+2log_4(x-1))+6log_4x#?

1 Answer
Carlos S.
Jan 18, 2018

Let #E = 1/2[log_4(x+1) + 2log_4(x-1)] + 6log_4x# 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^6#;

#E = 1/2log_4[(x+1)(x-1)^2] + log_4x^6#;

#E = log_4[(x+1)^(1/2)(x-1)] + log_4x^6#;

#E = log_4[(x+1)^(1/2)(x-1)x^6]#.

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