Question

find domain of the function given in the figure
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Answer 1

Answer:

(2/3 , 3]

Step-by-step explanation:

First we will check domain of log variables

For log x to be defined , x > 0 should hold

So

x - 2 > 0

x > 2              ---------------- eq. 1

3x - 1 > 0

x > 1/3         ------------------ eq.2

Now for sqrt (x) , x >= 0 should hold

So,

log₂(x-2) / log ₁/₂ (3x -1)  >= 0

[log₂(x-2)] / [(-) log₂(3x-1)] >= 0

log₂(x-2) / log₂(3x-1) <= 0

for log x < 0 , x < 1 should hold

so for fraction to be negative one of the num. and den. be negative and other positive and den. non zero

either (x-2) >= 1 & (3x-1) <1 or (x-2) <= 1 & (3x-1) >1

(x-2) >= 1 & (3x-1) <1

x >= 3 & x < 2/3  -> not possible

(x-2) <= 1 & (3x-1) >1

x <= 3 & x > 2/3

Common value is (2/3 , 3]    --------------- eq.3

Intersection of eq. 1 ,2 &3 is (2/3 ,3]

Hope it helps :-)