Linty Fresh
Junior Member
- Joined
- Sep 6, 2005
- Messages
- 58
Solve for x:
log2(x+1) - log4x=1
To solve it, I want to convert log base 4 to log base 2. Thus:
log4x=y
4^y=x
(2^2)^y=x
2^2y=x
2^y=sqrt(x)
Sooooo:
log2(x+1) - log2sqrt(x)=1
log2[(x+1)/sqrt(x)]=1
2^1=(x+1)/sqrt(x)
2=(x+1)/sqrt(x)
2*sqrt(x)=x+1
4x=x^2+2x+1
x^2-2x+1=0
(x-1)^2=0
x=1
Now according to the book, I got this one right, but then I tried solving using the method above for another problem and I got that wrong. Did I do this right?
log2(x+1) - log4x=1
To solve it, I want to convert log base 4 to log base 2. Thus:
log4x=y
4^y=x
(2^2)^y=x
2^2y=x
2^y=sqrt(x)
Sooooo:
log2(x+1) - log2sqrt(x)=1
log2[(x+1)/sqrt(x)]=1
2^1=(x+1)/sqrt(x)
2=(x+1)/sqrt(x)
2*sqrt(x)=x+1
4x=x^2+2x+1
x^2-2x+1=0
(x-1)^2=0
x=1
Now according to the book, I got this one right, but then I tried solving using the method above for another problem and I got that wrong. Did I do this right?