How do you find the domain and range of #y=log(2x-3)/(x-5) #?

Question

Eli Assouline · Accepted Answer

The domain is # x in (3/2,5)uu(5,+oo)#. The range is #y in RR#

The role is

#y=ln(2x-3)/(x-5)#

There are two things to think about for the domain:

#{(2x-3>0),(x-5!=0):}#

#=>#, #{(x>3/2),(x!=5):}#

Consequently,

The domain is # x in (3/2,5)uu(5,+oo)#

Calculate the following limits for the range.

#lim_(x->3/2)y=+oo#

#lim_(x->5^-)y=-oo#

#lim_(x->5^+)y=+oo#

#lim_(x->oo)y=0^+#

The range is #y in RR#

graph{ln(2x-3)/(x-5) [-16.77, -4.24, 7.01, 5.73]}

Eva Adamson · Answer

undefined To find the domain and range of $ y = \frac{\log(2x - 3)}{x - 5} $:

1. **Domain**: Set the denominator $ x - 5 $ not equal to zero and determine any restrictions on the argument of the logarithmic function.

For the denominator $ x - 5 \neq 0 $, so $ x \neq 5 $.
   
   For the argument of the logarithmic function $ 2x - 3 $, ensure that $ 2x - 3 > 0 $ since the logarithm of a non-positive number is undefined.
   
   Solve $ 2x - 3 > 0 $ for $ x $:
   $$ 2x - 3 > 0 $$
   $$ 2x > 3 $$
   $$ x > \frac{3}{2} $$

Therefore, the domain is $ x \in \left(\frac{3}{2}, 5\right) \cup (5, \infty) $.

2. **Range**: The range of $ y = \frac{\log(2x - 3)}{x - 5} $ is the set of all possible output values of $ y $.

As $ x $ approaches 5, $ \frac{\log(2x - 3)}{x - 5} $ approaches negative infinity.

As $ x $ approaches positive infinity, $ \frac{\log(2x - 3)}{x - 5} $ approaches 0.

Therefore, the range of the function is $ y \in (-\infty, 0) $.