Night reader wrote:Whit, (7^x)*49 in the numerator and 49 in the denominator are two different numbers in the mods
one is 7^2 from original source number, and the other is 49 - we don't know the original number
further to equate the powers one needs to have the same baseline number on both sides (left and right)
we have +ve 7 on the left and -ve 7 on the right in case when |7|=-7
So when I said can't be further simplified, it was meant that this expression breaks down to two expressions exactly:
7^(x+2)>|7|^2 AND |7|>7 {+ve} --> 7^(x+2)>7^2 This we can simplify and write as x+2>2
7^(x+2)>|7|^2 AND |7|>-7 {-ve} --> 7^(x+2)>-7^2 Here we can't write as x+2>2 because the baseline numbers are of the different values 7 and -7
OR in the more straightforward way
7^(x+2)/|7|^2 >1 --> for |7| -ve, 7^(x+2)<-7^2 AND for |7| +ve, 7^(x+2)>7^2 ... actually this would be the correct way, as we can not multiply inequality to the mod. What do you think?
49/49 will be 1 regardless of the underlying values used to create those 49s. To put this in a slightly different perspective - if I asked you for 81/81, you would not have to first ask if one was 9^2 and the other (-9)^2 or answer +1 or -1; the answer would simply be 1. If the numerator was made of (-9)(-9) and the denominator by (9)(9), the answer would only be 1 because the negatives will be paired and will cancel. We only have to worry about the mod or underlying sign of 7 if we are trying to pull out only one of the 7s in the denominator - but we are treating them as a whole. We can think about it in the following way:
(-7)*(-7) = (7)*(7)
(-7)^2 = 7^2
49 = 49
At every step of the expression we are equal. We would only need to be concerned with these "hidden" signs if we decided to remove one of the terms (i.e. take the square root). Because we do not have to do that to solve the problem asked in the post, we are home free.
In the step you noted here:
Night reader wrote:7^(x+2)>|7|^2 AND |7|>-7 {-ve} --> 7^(x+2)>-7^2 Here we can't write as x+2>2 because the baseline numbers are of the different values 7 and -7
Be careful, because I can rewrite (-7)^2 as (-1)^2 times (7)^2 and this is a legal move. And since (-1)^2 = 1, that term effectively disappears.
Again, 3 different methods will get you to the same solution, x>0, and only one of them utilizes "factoring" of the 49 in the denominator.
Whit
