Anyone here know Algebra?

[dleccord 5,000+ posts]

1. Find all real solutions to log_6(x)+log_6(x+1)+log_6(x+2)=1

2. Show that if sin2x + sin2y = 2sin(x + y) then either sinx = siny or cosx = cosy

Thanks!

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[dleccord 5,000+ posts]

sin2x+sin2y=2sin(x+y)

2sinxcosx+2sinycosy=2[sinxcosy+cosxsin...

sinxcosx+sinycosy= sinxcosy+cosxsiny

sinxcosx-sinxcosy+sinycosy-cosxsiny=0

sinx(cosx-cosy)-siny(-cosy+cosx)=0

sinx(cosx-cosy)-siny(cosx-cosy)=0

(sinx-siny)(cosx-cosy)=0

therefore sinx-siny=0 or cosx-cosy=0

sinx=siny or cosx=cosy QED

 

[Supergumby5000 5,000+ posts]

1) distribute

log6[(x)+(x+1)+(x+2)]=1

you should be able to get the rest of it, im too lazy to type

 

[dleccord 5,000+ posts]

log_6_[x+x+1+x+2]=1

6^{1}=[x+x+1+x+2]

6=[3x+3]

3x=3

x=1

Can someone check the work?

1) distribute
log6[(x)+(x+1)+(x+2)]=1

you should be able to get the rest of it, im too lazy to type

Thanks

 

[dleccord 5,000+ posts]

eh I made a mistake according to laws of log.

log_6_[(x(x+1)(x+2)]=1

log_6_[(x^3)+(3x^2)+(2x)]=1

6^1=[(x^3)+(3x^2)+(2x)]

[(x^3)+(3x^2)+(2x)]=6

[(x^3)+(3x^2)+(2x)+6]=0

[(x^2)(x+3)+2(x+3)]=0

(x^2)(x+3)=0

x=-3since x^2=-2 does not exist.