SOLUTION: HELP ME PLEASE: SOLVE LOGa(8x+5)=LOGa(4x+29)

Question 55179: HELP ME PLEASE:
SOLVE LOGa(8x+5)=LOGa(4x+29)
Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
SOLVE LOGa(8x+5)=LOGa(4x+29)
Because the bases are the same, we can simply equate:
8x+5=4x+29
-4x+8x+5=-4x+4x+29
4x+5=29
4x+5-5=29-5
4x=24
4x/4=24/6
x=6
check by substituting x=6 in the original equation. You have to do that with all log equations to avoid extraneous (false) solutions.
LOGa(8(6)+5)=LOGa(4(6)+29)
LOGa(48+5)=LOGa(24+29)
LOGa(53)=LOGa(53) We're right!!!
Happy Calculating!!!
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