Answer:
By graphing ; [tex]\log_2 (x-1) = \log_{12} (x-1)[/tex] only for x =2
Step-by-step explanation:
Solve: [tex]\log_2 (x-1) = \log_{12} (x-1)[/tex]
let [tex]y_1=\log_2 (x-1)[/tex] and [tex]y_2=\log_{12} (x-1)[/tex]
To find the x for which [tex]y_1= y_2[/tex]
A graph of these [tex]y_1=\log_2 (x-1)[/tex] and [tex]y_2=\log_{12} (x-1)[/tex] shows us that the graph intersect.
This implies that there is a single (x, y) value that satisfies both equations.
i.,e (2, 0)
Therefore, [tex]\log_2 (x-1) = \log_{12} (x-1)[/tex] only when x =2
You can see the graph as shown below:
