# 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?

Bit stuck on this one, please help.

Angus

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

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

if so it can be adjusted as,

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

remove the logs and proceed

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

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
$\displaystyle log_2\left(\frac{x}{2x-1}\right)= 0$
and, while it is impossible to take the logarithm of 0, the logarithm, to any base, of 1 is equal to 0.
$\displaystyle \frac{x}{2x- 1}= 1$ 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 $\displaystyle 2^{log_2(a)}= a$ ($\displaystyle log_2(x)$ and $\displaystyle 2^x$ are inverse functions) that tells you immediately that 2x- 1= x.

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

Bit stuck on this one, please help.

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 :-)

### log2(2x 1)-log2x=2

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