How do you solve logarithmic equations such as log(2x)= -1 or 3log(2)+log(x)= 3?
1 answer:
Remember some simle log rules
loga+logb=log(ab)
xloga=loga^x
means 
also, if no base is stated, assume base 10
so
log(2x)=-1 is base 10 and translates to
10^-1=2x
1/10=2x
divide both sides by 2 aka times 1/2
1/20=x
3log(2)+log(x)=3
log(2^3)+log(x)=3
log(8)+log(x)=3
log(8x)=3
base 10
10^3=8x
1000=8x
divide both sides by 8
x=125
loga+logb=log(ab)
xloga=loga^x
also, if no base is stated, assume base 10
so
log(2x)=-1 is base 10 and translates to
10^-1=2x
1/10=2x
divide both sides by 2 aka times 1/2
1/20=x
3log(2)+log(x)=3
log(2^3)+log(x)=3
log(8)+log(x)=3
log(8x)=3
base 10
10^3=8x
1000=8x
divide both sides by 8
x=125
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