How do you find the domain, x intercept and vertical asymptotes of #y=-log_3x+2#?

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Answer 1 · 2018-04-08T00:36:37

Domain: #(0, oo)# #x-#intercept: #(9,0)# Vertical asymptote: #x=0#

Recalling that the logarithmic function (in this case #log_3x#) does not exist for negative #x# or #x=0#, the domain is all #x>0,# or #(0, oo)#

The vertical asymptote will be at #x=0#, as the function is never truly defined for this value, but gets infinitely close to it.

The #x-#intercept can be found by setting the function equal to zero and solving for #x.#

#-log_3x+2=0#

#log_3x=2#

Recalling that #log_ab=ln(b)/ln(a)#, we rewrite as

#ln(x)/ln(3)=2#

#ln(x)=2ln3#

#2ln(3)=ln(3^2)=ln(9)#

So,

#ln(x)=ln(9)#

#x=9#

The #x-#intercept is #(9, 0)#