solve for x: log$2(x+1)+log$2(3x-5)=log$2(5x-3)+2

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solve for x: log$2(x+1)+log$2(3x-5)=log$2(5x-3)+2

Mathematics
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(x+1)(3x-5) = 4(5x-3)
okay......can you keep going?

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Other answers:

3x^2 -2x -5 = 20x -12 3x^2 - 22x +7 = 0
does the last line factor into (x-7)(3x-1)?
yes
x = 7 or 1/3 now substitute both these values into ur initial eqn to double check
okay and on the first line you wrote 4(5x-3).....how did you get the four? 2^2?
yes 2 can be written as log 4 to base 2
and then we use log a + log b =log (ab)
thanks you. thanks were i kept going wrong. The solutions both work. :)
no prob miss
can i ask one more thing? if you are solving for x and you take the square root to isolate x, do you include both the positive and negative answers?
yes
like lets say x^2 = k then x = k AND -k
in some special situations we can neglect the negative answer..bt generally it is essential to consider both
thanks you!

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