# Thread: how to solve log2 (x) = log2(2x-1)

1. ## how to solve log2 (x) = log2(2x-1)

Hi

I tried:
log2(x) - log2(2x-1) = 0

but then log2(0) not possible so could go nowhere with that.

I know 2^(log2(2x-1)) = 2^log2(x) but does that help me?

Angus

2. ## Re: how to solve log2 (x) = log2(2x-1)

I think what you've stated is $\log_2(x)-\log_2(2x-1)=0$

if so it can be adjusted as,

$\log_2(x)=\log_2(2x-1)$

remove the logs and proceed

3. ## Re: how to solve log2 (x) = log2(2x-1)

Writing $\log_{2} x = \log_{2} (2 x -1)$ and writing $x= 2x-1$ is exactly the same...

Kind regards

$\chi$ $\sigma$

4. ## Re: how to solve log2 (x) = log2(2x-1)

Originally Posted by angypangy
Hi

I tried:
log2(x) - log2(2x-1) = 0

but then log2(0) not possible so could go nowhere with that.
I don't see where "log2(0)" has anything to do with that. What you have is equivalent to
$log_2\left(\frac{x}{2x-1}\right)= 0$
and, while it is impossible to take the logarithm of 0, the logarithm, to any base, of 1 is equal to 0.
$\frac{x}{2x- 1}= 1$ which, of course, is the same as x= 2x- 1.

I know 2^(log2(2x-1)) = 2^log2(x) but does that help me?
Well, yes! Since $2^{log_2(a)}= a$ ( $log_2(x)$ and $2^x$ are inverse functions) that tells you immediately that 2x- 1= x.

Or, you could argue that since $log_2(x)$ is a one to one function, $log_2(a)= log_2(b)$ gives, immediately, a= b. That is, because $log_2(x)= log_2(2x-1)$, you must have x= 2x- 1.

Angus

5. ## Re: how to solve log2 (x) = log2(2x-1)

Loga(x) + Loga(y) = x+y, based on this rule, x= 2x-1. -x=-1, hence x=1.
substitute to check....
log2(1)= 0 & log2(2-1)= log2(1)=0

yip its correct :-)