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...

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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)` .