logarithms

[janekela]

Hola. Como estas?
Well I got this sum 4 homework and i wondered if u could plz tell me where i went wrong
log(4x)+log(x-1)=1
these logs are to the base 3
log4x+logx-log1=1
log(4*x*x)/1 =1
log4x=1
4x^2=3^1
4x^2*1/2=3^1*1/2
4x=3^1/2
4x=1.732
x=0.433

I also got this sum but i cant pass this stage
1+2log(x+1)=log(2x+1)+log(5x+8)
these logs r 2 the base 10
log10+2log(x+1)=log(2x+1)+log(5x+8)

can u plz help me

 

[tkhunny]

log(4x)+log(x-1)=1
these logs are to the base 3
log4x+logx-log1=1

Ack!! Very bad.

log(x-1) ≠ log(x) - log(1)

Never do that again.

log(4x)+log(x-1) = 1
log(4*x*(x-1)) = 1

Since they are Base 3 logs

4*x*(x-1) = 3

Now what?

 

[ChaoticLlama]

In your second line is where you went wrong!

There is no property such that log(a - b) = log(a) - log(b). However, that is exactly what you did where you split up log(x-1) into log(x) - log(1)

What you should have done is use a real () property of logs where
log(ab) = log(a) + log(b)

So...
log(4x) + log(x-1) = 1

log(4x² - 4x) = 1

raise both sides to the power of 3.

4x² - 4x = 3