Posted by anonymous on Tuesday, June 8, 2010 at 6:22pm.
Your equation up to
2x^2-x-6 = 0 is correct, but I factored it to get
so x = 2 or x = -3/2 which is inadmissable
LS = log5 5 + log51
= 1+0 = 1 = RS
so x = 2
Thank you :) One question, however. Where did you get LS = log[base5]5 + log[base5]1?
At the beginning, LS = log[base5](2x+2) + log[base5](x-1)
Sorry, it just confused me, it would help if you'd explain that.
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