How do I "check" this by substituting each value x=2 & x=4 into log5(x+3)+log5(x-1)...

Question

campbell_st · Answer

$$\log _{5} (2 + 3) = 1$$    and $$\log _{5} (2 -1) = 0$$

so when x = 2 the answer would be 1  for a base 5 log system

campbell_st · Answer

same process for x = 4

anonymous · Answer

so when u say same process for x=4 do i put 4 into the equation as well

campbell_st · Answer

yes... repeat the process for x = 4.... its log(4+3) + log (4-1)

anonymous · Answer

so i should come up with 1 and 0 again in order to come out right

campbell_st · Answer

no you have log 7 and log 3      The reason the 1st part was so easy in base 5 logs,    5 = 5^1 so then answer is 1 and 5^0 = 1  so the answer is 0...

anonymous · Answer

o ok so there is no substituting you just finish it out

anonymous · Answer

so when u have log7(4+3)=1 and log3(4-1)=1

campbell_st · Answer

well the only other suggestion is 

log(x+3) + log(x-1)    is the same as log{(x+3)(x-1)} using log laws.... so it could be log (7 x 3) = log 21

anonymous · Answer

ok???

campbell_st · Answer

lol... sorry... just they are funny numbers for base 5 logs.... change of base is an option if you need a number answer

anonymous · Answer

ok so earlier when i put log7(4+3)=1 & log3(4-1)=1 would that be right