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Find all real solutions to the equation
1 1
8 2
2
2
x x
x x
(Hint: The solution set has exactly 5 elements. If a,bεR, when will a
b
be equal to 1?)
Solution:
Let
1
2
x x a
---base
8 2
2
x x
b
---exponent
There are three possible ways that make the equation equal to 1.
In accord with the laws of indices/exponent, the equation is equal to 1 if:
1. In the expression given (a
b
), b =0 & a is any integer (except 0, because 0
0
is indeterminate)
Recall: x
0
=1
Finding the real solution
a. Equate 8 2
2
x x to 0, that is
0 8 2
2
x x
b. Factor the polynomial
2 4 x x
c. Equate both factors to 0 and find the values of x, that is
0 4 x 0 2 x
4 x 2 x
x = 4 , x = -2
2. In the expression given(a
b
), a is 1 & b is any integer
Recall: 1
n
= 1
Finding the real solution
2
a. Equate 1
2
x x to one, that is
1 1
2
x x
Transpose 1 to the left, thus becomes:
0 2
2
x x
b. Factor the Polynomial
2 1 x x
c. Equate both factors to 0 and find the values of x, that is
0 1 x 0 2 x
1 x 2 x
x = -1 , x = 2
3. In the expression given(a
b
), a is -1 & b is even integer
Recall: (-1)
n
= 1, where n is an even integer
Finding the real solution
a. Find the values of x for which 1 1
2
x x
1 1
2
x x
0
2
x x
By factoring: 0 1 x x
0 x ; 0 x
0 1 x ; 1 x
b. For each solution, let’s see if the exponent is an even integer by substituting the values of x
Let 0 x
8 2
2
x x ; 8 ) 0 ( 2 0
2
8 0 0 8 8 0 (EVEN)
Let 1 x
8 2
2
x x ; 8 ) 1 ( 2 1
2
8 2 1 9 10 1 (ODD)
Thus, the value of x that makes the exponent equal to an even integer is 0.
x = 0
TABLE OF SUMMARY
3
VALUE OF a VALUE OF b REAL SOLUTIONS (x) RESULT
any integer,
except 0
0 4 , -2 1
1 any integer -1 , 2 1
-1 even integer 0 1