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