Option c: x=6 is a true solution and x=-6 is an extraneous solution.
Explanation:
The equation is ![2 \log _{6} x=2](https://tex.z-dn.net/?f=2%20%5Clog%20_%7B6%7D%20x%3D2)
Dividing both sides of the equation by 2, we get,
![\frac{2 \log _{6}(x)}{2}=\frac{2}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B2%20%5Clog%20_%7B6%7D%28x%29%7D%7B2%7D%3D%5Cfrac%7B2%7D%7B2%7D)
Simplifying,
![\log _{6}(x)=1](https://tex.z-dn.net/?f=%5Clog%20_%7B6%7D%28x%29%3D1)
Since, we know by the logarithmic definition, if
, then ![b=a^{c}](https://tex.z-dn.net/?f=b%3Da%5E%7Bc%7D)
Using this definition, we have,
![x=6^{1}](https://tex.z-dn.net/?f=x%3D6%5E%7B1%7D)
Hence, ![x=6](https://tex.z-dn.net/?f=x%3D6)
Now, let us verify if
is the solution.
Substitute
in the equation
to see whether both sides of the equation are true.
We have,
![2 \log _{6}(6)](https://tex.z-dn.net/?f=2%20%5Clog%20_%7B6%7D%286%29)
Using the log rule, ![\log _{a}(a)=1](https://tex.z-dn.net/?f=%5Clog%20_%7Ba%7D%28a%29%3D1)
We have,
![2*1=2](https://tex.z-dn.net/?f=2%2A1%3D2)
Hence, both sides of the equation are equal.
Thus,
is the true solution.