`1/3log_5(12x)=2` Solve the equation. Check for extraneous solutions.

To evaluate the given equation `1/3log_5(12x)=2` , we may apply logarithm property: `n* log_b(x) = log_b(x^n)` .

`log_5((12x)^(1/3))=2`

Take the "log" on both sides to be able to apply the logarithm property: `a^(log_a(x))=x` .

`5^(log_5((12x)^(1/3)))=5^(2)`

`(12x)^(1/3)= 25`

Cubed both sides to cancel out the fractional exponent.

`((12x)^(1/3))^3= (25)^3`

`(12x)^(1/3*3)=15625`

`(12x)^(3/3)=15625`

`12x =15625`

Divide both sides by `12` .

`(12x)/12 =(15625)/12`

`x =(15625)/12`

Checking: Plug-in `x=(15625)/12` on `1/3log_5(12x)=2` 

`1/3log_5(12*(15625)/12)=?2`

`1/3log_5(15625)=?2`

`log_5(15625^(1/3))=?2`

`log_5(root(3)(15625))=?2`

`log_5(25)=?2`

`log_5(5^2)=?2`

`2log_5(5)=?2`

`2*1=?2`

`2=2...

To evaluate the given equation `1/3log_5(12x)=2` , we may apply logarithm property: `n* log_b(x) = log_b(x^n)` .

`log_5((12x)^(1/3))=2`

Take the "log" on both sides to be able to apply the logarithm property: `a^(log_a(x))=x` .

`5^(log_5((12x)^(1/3)))=5^(2)`

`(12x)^(1/3)= 25`

Cubed both sides to cancel out the fractional exponent.

`((12x)^(1/3))^3= (25)^3`

`(12x)^(1/3*3)=15625`

`(12x)^(3/3)=15625`

`12x =15625`

Divide both sides by `12` .

`(12x)/12 =(15625)/12`

`x =(15625)/12`

Checking: Plug-in `x=(15625)/12` on `1/3log_5(12x)=2` 

`1/3log_5(12*(15625)/12)=?2`

`1/3log_5(15625)=?2`

`log_5(15625^(1/3))=?2`

`log_5(root(3)(15625))=?2`

`log_5(25)=?2`

`log_5(5^2)=?2`

`2log_5(5)=?2`

`2*1=?2`

`2=2 `  TRUE

There is no extraneous solution. The `x=(15625)/12` is a real solution for the given equation `1/3log_5(12x)=2` .