To evaluate the given equation 36^(5x+2)=(1/6)^(11-x) , we may apply 36=6^2 and 1/6=6^(-1) . The equation becomes:
(6^2)^(5x+2)=(6^(-1))^(11-x)
Apply Law of Exponents: (x^n)^m = x^(n*m) .
6^(2*(5x+2))=6^((-1)*(11-x))
6^(10x+4)=6^(-11+x)
Apply the theorem: If b^x=b^y then x=y , we get:
10x+4=-11+x
Subtract x from both sides of the equation.
10x+4-x=-11+x-x
9x+4=-11
Subtract 4 from both sides of the equation.
9x+4-4=-11-4
9x=-15
Divide both sides by 9 .
9x/9=-15/9
x=-15/9
Simplify.
x=-5/3
Checking: Plug-in x=-5/3 on 36^(5x+2)=(1/6)^(11-x) .
36^(5(-5/3)+2)=?(1/6)^(11-(-5/3))
36^(-25/3+2)=?(1/6)^(11+5/3)
36^(-25/3+6/3)=?(1/6)^(33/3+5/3)
36^(-19/3)=?(1/6)^(38/3)
(6^2)^(-19/3)=?(6^(-1))^(38/3)
6^(2*(-19/3))=?6^((-1)*38/3)
6^(-38/3)=6^(-38/3) TRUE
Final answer:
There is no extraneous solution. The x=-5/3 is the real exact solution of the equation 36^(5x+2)=(1/6)^(11-x) .
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