Algebra 2, please help!!!!!

log₅(x) + log₅(2x + 3)  =  1

Since logs are added when the original problem was multiplied, going backwards we have:

log₅[ x · (2x + 3) ]  =  1

log₅( 2x² + 3x )  =  1

Writing the log in exponential notation:

2x² + 3x  =  5¹

--->   2x² + 3x  =  5     --->     2x² + 3x - 5  =  0     --->     (2x + 5)(x - 1)  =  0

                                                                         --->     Either  2x + 5  =  0     or     x - 1  = 0

                                                                         --->     Either     x  =  - 5/2     or     x  =  1

Since  log₅(x)  is undefined if x is a negative number, the value  -5/2  must be ignored; thus the answer is  x  =  1