# Can someone just help explain a clear defination of superlog?

In summary, the superlog and superroot equations can be solved by using a computer program to find an approximate solution for x, as long as certain conditions are met such as x being greater than 0 and log_c(x) being in the range of 0 to 1. This can be done without needing to know complex analysis or other advanced mathematical concepts.
I was trying to understand superlog and superroot but I get only 3/4 of them. Can anyone just explain, in a non-textbook way, such that:
I can understand without any post-calc knowledge
http://en.wikipedia.org/wiki/Superlog
-------------------------------------------------
or just explain:
c=x^x
how to find x?
or c=x^x^x^x^x^x^x....
thanks. just to say again that I am not even doing calc in school, but I am doing all maths myself. So will you explain without using any complex analysis and other things?

I was trying to understand superlog and superroot but I get only 3/4 of them. Can anyone just explain, in a non-textbook way, such that:
I can understand without any post-calc knowledge
http://en.wikipedia.org/wiki/Superlog
-------------------------------------------------
or just explain:
c=x^x
how to find x?
or c=x^x^x^x^x^x^x....
thanks. just to say again that I am not even doing calc in school, but I am doing all maths myself. So will you explain without using any complex analysis and other things?

Hey n_kelthuzad and welcome to the forums.

The reason why we need complex analysis is because the solution x could actually be negative. When x is negative then we are looking at most likely a complex number solution depending on what x actually is.

So let's for the moment assume that x > 0. We only consider a finite number of exponentiations: Let's look at two first.

c = x^x^x. Assume x > 0 and c > 0 (It has to be otherwise it won't work).

log_c(c) = log_c(x^x^x) = 1

But log_c(x^x^x) = x^xlog_c(x) = 1.

This meanx x^x = 1/log_c(x).

Now let's do the same thing again:

log_c(x^x) = log_c(1/log_c(x)) = - log_c(log_c(x)) since log(1) = 0 and log(a/b) = log(a) - log(b).

But log_c(x^x) = xlog_c(x) which means:

xlog_c(x) = -log_c(log_c(x)) which means

x = -log_c(log_c(x)) / log_c(x)

To solve this, you need to use a computer for the general case. Now log_c(x) has to be greater than 0 otherwise you won't get a solution. This means that x > c for the real case (remember no complex numbers).

Since log_c(x) > 0 and since x > 0 this means log_c(log_c(x)) < 0 since we have a minus sign but this means log_c(log_c(x)) < 0 which means log_c(x) is in 0 < x < 1.

If those conditions are satisfied we use a computer program which takes in a formula and finds what is called the root of the equation: in this particular example - we have x + log_c(log_c(x))/log_c(x) = 0 and we have to solve for x. The computer will then do its magic and get an approximate answer for x.

Also just in case you are wondering log_c(x) = ln(x)/ln(c) where ln is the natural logarithm function which has a known formula.

## 1. What is the definition of superlog?

Superlog is a term used in computer science and mathematics to describe a logic system that is highly efficient and optimized for use in electronic circuits.

## 2. How is superlog different from traditional logic systems?

Superlog is different from traditional logic systems because it uses a different set of rules and symbols to represent logical operations. It also allows for more complex and efficient operations to be performed, making it more suitable for use in electronic circuits.

## 3. What are some examples of superlog being used?

Superlog is commonly used in the design and verification of electronic circuits, such as in microprocessors, ASICs, and FPGAs. It is also used in software development for high-performance systems.

## 4. How does superlog improve efficiency in electronic circuits?

Superlog improves efficiency in electronic circuits by allowing for more complex operations to be performed with fewer logical gates. This results in smaller and faster circuits, reducing power consumption and improving overall performance.

## 5. Are there any drawbacks to using superlog?

One potential drawback of using superlog is that it requires specialized knowledge and skills to design and implement. It may also be more difficult to debug in comparison to traditional logic systems. Additionally, not all electronic design tools support superlog, so it may not be a viable option for all projects.

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