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Mathematics 14 Online
OpenStudy (sphott51):

Solve log(3x+5) =0

OpenStudy (freckles):

0=log(1)

OpenStudy (freckles):

again if log(a)=log(b) then a=b

OpenStudy (freckles):

replace 0 with log(1) as suggested in my first post and then use my second post to solve a linear equation

OpenStudy (freckles):

untrue |dw:1447125672052:dw|

OpenStudy (xapproachesinfinity):

that is just a disguized linear equation 3x+5=1 like @freckles mentioned log1=0 so what inside log must equal to 1 for that to happen

OpenStudy (solomonzelman):

I was thinking of somthing else..

OpenStudy (xapproachesinfinity):

lol @SolomonZelman sometimes i do such mistakes hahaha

OpenStudy (xapproachesinfinity):

the other i had cal iii exam i did a severe mistake like that hhhhhh i know the answer yet i went for the wrong thing lol

OpenStudy (xapproachesinfinity):

the other day*

OpenStudy (solomonzelman):

I overthought and started to talk nonsense

OpenStudy (sphott51):

Okay so what do you mean? Like log3x +log5 =0..?

OpenStudy (freckles):

no @SPHott51

OpenStudy (xapproachesinfinity):

nn dear that is not true at all about logs

OpenStudy (freckles):

log(1) is 0 so replace 0 with log(1) log(3x+5)=log(1) implies 3x+5=1 solve 3x+5=1 for x

OpenStudy (xapproachesinfinity):

i wrote the equation above 3x+5=1

OpenStudy (freckles):

maybe he got confused from all the other stuff being posted about the domain versus range thing @xapproachesinfinity

OpenStudy (xapproachesinfinity):

i guess :) hopefully it is clearer now

OpenStudy (freckles):

though that bit is also important to know :p

OpenStudy (sphott51):

AGHH, I always make things harder for myself

OpenStudy (xapproachesinfinity):

think of log in association with exponential since they are inverse of each other to make your life easier

OpenStudy (xapproachesinfinity):

for instance we know that (anything)^0=1 so log1=0 it is important to understand this bond

OpenStudy (xapproachesinfinity):

for porperties of log(ab)=loga +logb for instance came from the fact that 10^(a+b)=10^a 10^b (im using log base 10 here, but it really does not matter for properties of logs)

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