log8 + log8 (x+2)=1 . Find x

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log8 + log8 (x+2)=1 . Find x

Mathematics
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think you are missing something. i am guessing this is \[log_8(x)+log_8(x+2)=1\]
yes your right
ok in that case straight forward. step one is combine the logs on the right into a single logarithm using \[log(ab)=log(a)+log(b)\] backwards

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you get \[log_8(x)+log_8(x+2)=log_8(x(x+2))\]
so \[log_8(x(x+2))=1\] in equivalent exponential form this mean \[x(x+2)=8^1=8\]
now you have a quadratic equation to solve: \[x(x+2)=8\] \[x^2+2x=8\] \[x^2+2x-8=0\] factor \[(x+4)(x-2)=0\] \[x=-4\] or \[x=2\] but now you have to be careful
because one answer is negative and you cannot take the log of a negative number. so the answer is not -4 or 2, the answer is just 2
Thank you so much :-D <3
welcome hope the steps were clear.

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