Please tell me if this is right... Solve the logarithmic equation: log_4(x-18) + log_4 (x-3) = 2 log_4((x-18)(x-3)=4^2 x^2-21x+54=16 x^2-21x+38=0 (x-2)(x-19)=0 x=2, x=19 <--- Answer

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Please tell me if this is right... Solve the logarithmic equation: log_4(x-18) + log_4 (x-3) = 2 log_4((x-18)(x-3)=4^2 x^2-21x+54=16 x^2-21x+38=0 (x-2)(x-19)=0 x=2, x=19 <--- Answer

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your final answer is correct but the 2nd step will be log_4((x-18)(x-3)=2 then 3rd step x^2-21x+54=16 is correct
or your 2nd step could be log_4((x-18)(x-3)=log_4 (4^2)
yea I did that but I skipped the step on here. I wrote that on my paper.

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then you are correct
oh, wait i just realized, x=2 cannot be the answer
because that would make x-3 as negative then log (x-3) will not be defined @Fakshon

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