# Thread: inverse of 0.5^2

1. ## inverse of 0.5^2

What is the inverse function of 0.5^2?

2. The question seems strange. The inverse of a number is a number, and the inverse of a function is a function (or, more generally, a relation). Here you seem to claim that the inverse of a number is a function.

3. As emakarov said, in order to have an inverse function, you have to have a function! $\displaystyle 0.5^2= 0.25$ is a number, not a function. If you mean $\displaystyle f(x)= x^2$, for x non-negative, then its inverse is $\displaystyle f^{-1}(x)= \sqrt{x}$. And, of course, $\displaystyle f^{-1}(0.25)= 0.5$. As long as f has an inverse, if $\displaystyle f(a)= b$, then $\displaystyle f^{-1}(b)= a$.

4. I meant to write 0.5^x. I know the inverse of for instance 2^x but not for values less than 1. Also I don't know how to find the inverse function of 2^-x which has a similar graph to 0.5^x.

5. Originally Posted by Stuck Man
I meant to write 0.5^x. I know the inverse of for instance 2^x but not for values less than 1. Also I don't know how to find the inverse function of 2^-x which has a similar graph to 0.5^x.
Well $\displaystyle (0.5)^x=2^{-x}$ so the inverse is $\displaystyle -\log_2(x)$

6. That is not it.

7. Originally Posted by Stuck Man
That is not it.
Why do you say that?

If $\displaystyle f(x)=(0.5)^x~\&~g(x)=-\log_2(x)$ then $\displaystyle f(g(x))=g(f(x))=x$.

8. The graph does not look right.

9. Try $\displaystyle g(x)=-\dfrac{\ln(x)}{\ln(2)}$

10. I was using that since my graph program does not support multible bases. I still think it doesn't look right. There is not any image of the negative values.

11. I tried it and it looks right to me... Remember that the graph of the inverse function will look like it's been reflected along the x=y line. This doesn't mean there won't be any overlap of the functions, and it doesn't mean the functions will be in opposing quadrants, which I think is what you're looking for.

Think about it. If f(x) returns y, then g(y) returns x if g(x) and f(x) are inverse functions. In this case, where $\displaystyle f(x) = y = 2^{(-x)}$, you can have any real x you want, but y will always be positive, no matter what your x value is. So basically, the function f(x) in this case takes ANY x value, and returns a positive y value. What does that mean about its inverse function? Well... that means that its inverse function will take any positive y value, and return some real x value (no restrictions). Since you're taking the inverse function of x, that means in this case your x will always be positive and your y will span across the real numbers.

Look at your graph again. Does it look like a reflection about the line y=x?

12. Originally Posted by Stuck Man
There is not any image of the negative values.
Think about it: $\displaystyle (0.5)^x>0$ for all $\displaystyle x$.

13. The logarithm function can only be used with positive values. The domain of the inverse function would be negative real values < 0.

14. I was expecting a reflection on the y=-x line. I'll look at it again now. Plato, I meant negative x values.

15. No. For values of x < 0, the domain will be > 0.

EDIT: Sorry, I meant for x being a fraction.

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