Question

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

Answer 1

think you are missing something. i am guessing this is
\[log_8(x)+log_8(x+2)=1\]

Answer 2

yes your right

Answer 3

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

Answer 4

you get
\[log_8(x)+log_8(x+2)=log_8(x(x+2))\]

Answer 5

so \[log_8(x(x+2))=1\] in equivalent exponential form this mean
\[x(x+2)=8^1=8\]

Answer 6

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

Answer 7

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

Answer 8

Thank you so much :-D <3

Answer 9

welcome hope the steps were clear.