2 Answers

Radhakrishnan · Answer 1 · 2018-02-26T11:38:03+0000

Answer:

By graphing ;
$\log_2 (x-1) = \log_(12) (x-1)$ only for x =2

Explanation:

Solve:
$\log_2 (x-1) = \log_(12) (x-1)$

let
$y_1=\log_2 (x-1)$ and
$y_2=\log_(12) (x-1)$

To find the x for which

A graph of these
$y_1=\log_2 (x-1)$ and
$y_2=\log_(12) (x-1)$ 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,
$\log_2 (x-1) = \log_(12) (x-1)$ only when x =2

You can see the graph as shown below:

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