(p^q)-(q^r)= (p + q)^r-q where 1 <q <r<p <10 then the value of p+q+r is:
Answers
Answer:
Since, p is true and q is false.
∴p→q has truth value F.
Statement r has truth value T.
∴(p→q)∧r has truth value F.
Also, (p→q)∧∼r has truth value F.
Now, p∧q has truth value F and p∨r has truth value T.
∴(p∧q)∧(p∨q) has truth value F.
As, p∧r has truth value T.
Therefore, q→(p∧r) has truth value T.
Answer:
Step-by-step explanation:
If we have a look at a couple of examples we see that they turn out to be in A.P. So let's prove this.
If x, y, z are in GP then we can write x = y/r, z = ry for some nonzero r. Then we have
(y/r)^p = y^q = (ry)^b
Taking logs gives
p (log y - log r) = q log y = b (log y + log r)
From the first pair, solve for log r:
p log y - p log r = q log y
=> log r = (p log y - q log y) / p
= (1 - q/p) log y
Do the same from the second pair:
q log y = b log y + b log r
=> log r = (q log y - b log y) / b
= (q/b - 1) log y
Equate these two expressions for log r:
(1 - q/p) log y = (q/b - 1) log y
Now if log y = 0 then y = 1, so x = 1/r and z = r. This gives 1/r^p = 1 = r^b, hence r^p = r^b = 1, which only works if p = b = 0 or r = 1. In the former case the question is not well defined since 1/p and 1/b do not exist; in the latter we have x = y = z = 1 and p, q, and b can be any numbers so clearly no relation must hold.
However, if we put aside the vexing case y = 1, we get from the previous equation
(1 - q/p) = (q/b - 1)
and dividing by q gives
1/q - 1/p = 1/b - 1/q
which means that 1/p, 1/q, 1/b are in A.P.
So if we exclude the case y = 1, then 1/p, 1/q, 1/b are in A.P.
We can tighten this up a little more:
If x, y, and z are not all equal to 1, and p and b are not zero, then 1/p, 1/q, 1/b are in A.P.
However as stated None is the correct answer, because in the case x = y = z = 1 we can have p, q, b any numbers and so no relationship need exist between 1/p, 1/q, and 1/b. (And in the case x = 1/r, y = 1, z = 1/r there is a solution with p = b = 0 and q any number, in which case 1/p and 1/b do not exist.)