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`36^(5x+2)=(1/6)^(11-x)` Solve the equation.

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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