Question 29634: 81^x=9^(x^2-3)
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! 81^x=9^(x^2-3)
[(9)^2]^x = 9^(x^2-3)----(1)
(9)^(2x) = 9^(x^2-3)
This implies
(2x) = (x^2-3) (bases same implies powers equal)
0 = x^2-2x-3
That is x^2-2x-3 = 0
(x-3)(x+1) = 0
(x-3)= 0 gives x=3
(x+1) = 0 gives x = -1
Answer: x = 3 and x = -1
Verification: x=3 in (1)
LHS = [(9)^2]^x = [(9)^2]^3 = [(9)^6]
RHS = 9^(x^2-3)= 9^(3^2-3)= [(9)^6]=LHS
x=-1 in (1)
LHS = [(9)^2]^(-1) = [(9)^(-2)]
RHS = 9^[(-1)^2-3)= 9^(1-3)= [(9)^(-2)]=LHS
Therefore our answers are correct
Note: Given x^2-3x+x-3 = 0 how to factorise the LHS expression.
[splitting the middle term into two parts so that their sum is the middle term and their product is the product of the square term and the constant term. Here (-2x) = (-3x)+(x) and (-3x)X(x) = -3x^2 = (x^2)X(-3)]
(x^2-3x)+(x-3) = 0
x(x-3)+1(x-3)=0
xp+p= 0 where p = (x-3)
p(x+1) = 0
(x-3)(x+1) = 0
(x-3)= 0 gives x=3
(x+1) = 0 gives x = -1
Answer: x = 3 and x = -1
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