How do you find the domain, x intercept and vertical asymptotes of #f(x)=ln(x-1)#?
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Domain: #(1,infty)#
#x#-intercept: #x=2#
Vertical asymptote: #x=1#
All of the questions can be answered easily by realising that this function is just a translated version of the original logarithm.
In fact, it is true for any function #f# that the graph of #f(x)# and #f(x+k)# only differ for a horizontal translation, #k# units to the left if #k>0#, to the right otherwise.
So, in your case, the graph of #ln(x-1)# is identical to the graph of #ln(x)#, except it is translated one unit right.
So, since the domain of #ln(x)# is #(0,\infty)#, the domain of #ln(x-1)# will be #(1,infty)#. This makes sense, since the logarithm only accepts positive inputs, and #x-1# is positive only when #x>1#.
For the same reason, since #ln(x)=0# for #x=1#, then #ln(x-1)=0# for #x=2#, which is its #x#-intercept.
Finally, since #ln(x)# has a vertical asymptote at #x=0#, #ln(x-1)# will have a vertical asymptote at #x=1#.
Here's a graph for a visualization of all the computations:
graph{ln(x-1) [-1.584, 8.416, -2.08, 2.92]}
