I am trying to find the domain of $f(x)=\ln(3x-4)$. I cannot find out how to get the domain. but I did manage to get the vertical asymptote which is $x=4/3$.
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$\begingroup$ Use that the domain of $\log(z)$ is $z>0$. Now just plug in '$z$' for your case and read off the answer. $\endgroup$– WintherJun 1, 2014 at 20:44
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1$\begingroup$ so the answer is x>4/3 so I already had the answer and needed to change the '=' to '>' for domain $\endgroup$– DrewJun 1, 2014 at 21:14
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2 Answers
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So the domain of a function is the set of all $x$ for which $f(x)$ is defined. The trick here is remembering that the basic natural log function $f(x) = \ln(x)$ is only defined for $x > 0$ (it is the inverse function of $y = e^x$).
Given this, your function can only take inputs when the argument of the natural log is positive. Therefore, the domain is all $x$ such that $3x - 4 > 0$.
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The function $\ln(\text{something})$ is defined only and only if that $\text{something}$ is strictly positive, that is $\gt 0.$ (Why?) So here all what you have to do is to recognize that your $\text{something}$ is just the expression $3x-4$. So to know where the function we're discussing is defined, solve the inequality $3x-4\gt0$.