I'm not sure where you learned this, but there is a lot of extraneous math in your solution. All the "1^ -1" is completely unnecessary and confusing, although it does somehow give the correct solution. Here's a simpler way to do it:
256x-2 = 2-x-7 | 256 is equal to 28, so you can make the bases the same
28x-16=2-x-7 | Now, you can take away the bases and solve algebraically. If you perform log2 on both sides, you're left with just the exponents, so I'm just not showing a step.
8x-16=-x-7 | Next, you solve algebraically.
9x=9 | Divide both sides by nine.
x=1 | You have your answer!
Now, let's plug it back in to check if it works.
256x-2 = 2-x-7 | Replace x with our answer, 1
2561-2=2-1-7 | Change 256 to 28
28-16=2-1-7 | Simplify
It works! It may look hard at first, but once you try it out it's much easier and saves you the extra steps.
TLDR; Match your bases, take the log and solve algebraically.
50%. Since the outcomes of the previous flips don't affect the current flip, it merely depends on the number of flips and the odds of getting heads. Unless you want to count the miniscule chance of getting neither heads nor tails, where the coin lands on it's side and you slowly descend into madness as all math, physics, and probability start to fall apart, it's a pure 50% chance that you'll get heads. There is also an equal number of odd numbers between 1 and 2016, so it's not skewed odds. Thus, your answer is 50%.
So, the easiest way to solve this is to work backwards. Ed ends up with a grand total of $150. Before that, she earned an extra $50, so we subtract that from $150 leaving us with $100. If she spent half her paycheck and ended up with $100, then we can multiply that $100 by 2 giving us her initial weekly paycheck: $200
Here's the work in normal form:
Web2.0Calc is a site dedicated to helping people understand and appreciate all forms of mathematics. The forums are a place for people to ask questions about parts of math they don't understand, whether it's a specific question or a short explanation of a whole part of math. If you have any other questions, feel free to ask!
Happy to welcome you to the family,
It depends on the original format of the question. If it was originally written as 2*log(x+2) then no, you can't treat the two like a normal interger until you get rid of the logarithm. If, however, it was written 2*log(x)+2, then the two is not a part of the logarithm and can be treated as a normal interger because it is one. Once you determine the right format, it's merely a matter of stacking the logs and solving algebraically.
Hope this helped!