Answer :

Ilyes39
[tex]\log(5x-5)=\log(2x+3)+\log15\\ \log(5x-5)=\log[15(2x+3) ], \text{ because } \log(ab)=\log a+\log b\\ \log(5x-5)=\log(30x+45)\\ 5x-5=30x+45, \text{ because } (\log x=\log y\rightarrow x=y.)\\ -25x=50\\ \text{Then } x=-2.[/tex]

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