evaluate x if log2 (1+x) + log2 (5-x) - log2 (x-2)=3

Question

anonymous · Answer

$$ \log_2 (1+x) + \log_2 (5-x) - \log_2 (x-2)=3$$$$ \log_2[~~ (-x^2+4x+5) \div (x-2)~~]=3$$$$ \log_2[~~ (-x^2+4x+5) \div (x-2)~~]=3~\log_22$$$$ \log_2[~~ (-x^2+4x+5) \div (x-2)~~]=\log_28$$$$ (-x^2+4x+5) \div (x-2)=8$$$$ -x^2+4x+5=8x-16$$$$ -x^2-4x+21=0$$

anonymous · Answer

$$ -x^2-4x+21=0$$$$x^2+4x-21=0$$$$(x+7)(x-3)=0$$

anonymous · Answer

thanks you very much , but now i have another question, can you help me?
log5 (x)= 4logx (5) , evaluate x.    is it i have to change both of the log into same base?

mathmate · Answer

@kY_Tan 
To complete the solution, you wukk need to validate each of the solutions.  Whenever you're dealing with log functions, you need to verify that the solution does not give a negative argument for the log.

mathmate · Answer

*will

anonymous · Answer

@mathmate thanks

mathmate · Answer

@kY_Tan you're welcome! :)