Answer:
<h2>
![{f}^{ - 1} (x) = \frac{ln(x)}{ln(2)}](https://tex.z-dn.net/?f=%20%7Bf%7D%5E%7B%20-%201%7D%20%28x%29%20%3D%20%20%5Cfrac%7Bln%28x%29%7D%7Bln%282%29%7D%20)
</h2>
Step-by-step explanation:
![f(x)=2^x](https://tex.z-dn.net/?f=%20f%28x%29%3D2%5Ex)
![y = {2}^{x}](https://tex.z-dn.net/?f=y%20%3D%20%20%7B2%7D%5E%7Bx%7D%20)
![x = {2}^{y}](https://tex.z-dn.net/?f=x%20%3D%20%20%7B2%7D%5E%7By%7D%20)
![{2}^{y} = x](https://tex.z-dn.net/?f=%20%7B2%7D%5E%7By%7D%20%20%3D%20x)
![ln( {2}^{y} )=ln(x)](https://tex.z-dn.net/?f=ln%28%20%7B2%7D%5E%7By%7D%20%29%3Dln%28x%29)
![yln(2)=ln(x)](https://tex.z-dn.net/?f=yln%282%29%3Dln%28x%29)
<h3>
![\frac{yln(2)}{ln(2)} = \frac{ln(x)}{ln(2)}](https://tex.z-dn.net/?f=%20%5Cfrac%7Byln%282%29%7D%7Bln%282%29%7D%20%20%3D%20%20%5Cfrac%7Bln%28x%29%7D%7Bln%282%29%7D%20)
</h3><h3>
![y = \frac{ln(x)}{ln(2)}](https://tex.z-dn.net/?f=y%20%3D%20%20%5Cfrac%7Bln%28x%29%7D%7Bln%282%29%7D%20)
</h3><h3>
![{f}^{ - 1} (x) = \frac{ln(x)}{ln(2)}](https://tex.z-dn.net/?f=%20%7Bf%7D%5E%7B%20-%201%7D%20%28x%29%20%3D%20%20%5Cfrac%7Bln%28x%29%7D%7Bln%282%29%7D%20)
</h3><h3>Hope it is helpful...</h3>
I would say 504 basing it off of converting 2/3 into 4/6 and used sets of 84 to each be 1/6
Answer:
Yes
Step-by-step explanation:
Given that In my data set of 10 exam scores, the mean turned out to be the score of the person with the third highest grade.
No two people got the same score.
Let the scores be a,b,c,d,e in ascending order where no two scores are equal
If a+b=d+e =2c then we can have c as the mean of the scores of 5 persons
This is because
Total sum = a+b+c+d+e = (a+b)+c+(d+e)
= 2c+c+2c=5c
Average= 5c/5 = c
It makes sense and there are chances as long as the above condition is satisfied.
Answer:
They are very different because hexagons are amazing and squares are boring.
Step-by-step explanation:
900,000 would be the nearest hundred thousand
sorry if it's wrong, but I did try