Did I do this logarithmic function right?

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?
 
It might help if you posted that other problem, the one you're having trouble with. Right now, we have no idea what you might have done on it, other than that some of the methodology is similar to what you've displayed here.

By the way, to convert log<sub>4</sub>(x) to something in log-base-2, just use the change-of-base formula:

. . . . .log<sub>b</sub>(x) = log<sub>a</sub>(x)/log<sub>a</sub>(b)

. . . . .log<sub>4</sub>(x) = log<sub>2</sub>(x)/log<sub>2</sub>(4) = log<sub>2</sub>(x)/2

Eliz.
 
Hi, Stapel. Thanks so much for your help. Actually, I would rather just work through this and try to do the other problem on my own once I find out where I went wrong.

As far as I can tell, I made my mistake here:

log4x=y
4^y=x
(2^2)^y=x
2^2y=x
2^y=sqrt(x)

So converted, it should read log2(x)/2 rather than log2sqrt(x). Why is that? Sorry, but I'm a bit new to logarithms. Thanks again for your help.
 
Either way, it means the same thing, since sqrt(x) = x<sup>1/2</sup> and log<sub>b</sub>(x<sup>n</sup>) = n×log<sub>b</sub>(x).

Eliz.
 
Yeesh, of COURSE!! OK, let me work on that other math problem, and if I'm having trouble again, I'll post it in a new thread, thanks.
 
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