Question

What is the solution of log2x + 727 = 3? (Little 2x and little 7 before the 27)

Answer 1

$D:2x+7 \ \textgreater \ 0\ and\ 2x+7\neq1\\2x \ \textgreater \ -7\ and\ 2x\neq-6\\x \ \textgreater \ -3.5\ and\ x\neq-3\\\\\log_{(2x+7)}27=3\iff(2x+7)^3=27\\\\(2x+7)^3=3^3\iff2x+7=3\ \ \ |subtract\ 7\ from\ both\ sides\\\\2x=-4\ \ \ |divide\ both\ sides\ by\ 2\\\\\boxed{x=-2}\in D$

Answer 2

Remember
$log_{n}x=y$ means
y is the power to which n must be raised to obtain x
aka
n^y=x

also
if a^n=b^n where n=n, then a=b


$log_{2x+7}27=3$
translate
(2x+7)^3=27
see if we can match the bases
(2x+7)^3=3^3
therefor
2x+7=3
minus 7 
2x=-4
divide 2
x=-2