How do you solve #log (4p-2)=log(-5p+5)#?

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Answer 1 · 2018-04-05T14:37:15

#color(blue)(7/9)#

If:

#log(4p-2)=log(-5p+5)=>4p-2=-5p+5#

#4p-2=-5p+5#

#9p=7#

#p=7/9#

Check:

#log(4(7/9)-2)=log(-5(7/9)+5)#

#log(10/9)=log(10/9)#