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 2log_2(x+5) = log_2(x-9) + log_2(x+53)+1

 

Replace 1 by log_2(2)

2log_2(x+5) = log_2(x-9) + log_2(x+53) + log_2(2)

 

loga + logb = logab, so

2log_2(x+5) = log_2( (x-9)*(x+53)*2)

 

alogb = log(b^a), so

 

log_2( (x+5)² ) = log_2( (x-9)*(x+53)*2)

 

Hence

(x+5)² = 2(x-9)(x+53)

 

Can you take it from here?