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Granted that this is algebra, but logic can give a far more simplistic, yet still utterly correct, answer.
You did not specify that the value of a had to be numeric, therefore, the value of a is a . After all, if 0=0, 1=1, 2=2, etc, then a = a
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[ (9^0+8^0)/2 ] = a + 1
Find the numeric value of a to find, on a scale of 1-10, how much I care. :)
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Clever, my friend, truly clever. ;) I would change it to "Find the numeric value of a " to stop people like me.
Any number raised to power 0 is 1. So, ( 1+1)/2 = a + 1.. therefore, a = 0
9 ( 9 9 − 9 8 ) + 9 9 ⇒ a = 9 9 ( 9 − 1 ) + 9 9 = 9 9 ( 8 + 1 ) = 9 1 0 = 1 0
9 ( 9 9 − 9 8 ) + 9 9 = 9 a 9 1 0 − 9 9 + 9 9 = 9 a 9 1 0 = 9 a a = 1 0
9(9^9-9^8)+9^9=9^a =9^10-9^9+9^9=9^10=9^a =a=10
9(9^9-9^8)+9=9^9(9-1)+9^9=(9^9)(8+1)=(9^9)9=9^(9+1)=9^10 thus a=10
9((9^9)-(9^8))+9^9= 9(9^1)+9^9= 9x9+9^9= 9^11
Basically just look at the exponent to left to right And think of this equation 9-8+9=10
9*9^8(9^1-1)+9^9=9^a... as bases are same then we can add the powers 9^9(8)+9^9=9^a 9^9(8+1)=9^a.. 9^9(9)=9^a... 9^10=9^a. as bases are equal so their exponents are also equal. a=10 is answer
9(9^9 - 9^8) + 9^9 = 9^a
9^10 - 9^9 + 9^9 = 9^a
9^10 = 9^a
10=a
9 x ( 9^9 - 9^8 ) + 9^9 = 9^9 (9-1) + 9^9 = 9^9( 9 - 1 + 1 ) = 9^9 ( 9 ) = 9^10 :D
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9 1 0 − 9 9 + 9 9 = 9 a ⇒ 9 1 0 = 9 a ⇒ a = 1 0