This is related to Joe's answer, but I want to emphasize something else. The main point I want to make is that if $n^\text{th}$ roots are intuitive, then so are logarithms.
If you want to solve the equation $x^5=11$, you take a fifth root: $x=\sqrt[5]{11}$. But what does $\sqrt[5]{11}$ mean? Well, it means the unique real number $x$ such that $x^5=11$, so this alone doesn't tell you much. To get somewhat of an intuitive feel for it, you can look at the curve $y=x^5$, notice that it is always increasing, and that it crosses $y=11$ somewhere between $1$ and $2$ (because $1^5=1$ and $2^5=32$). But ultimately the definition of $\sqrt[5]{\ }$ relies on the notion of inverting the more familiar operation of multiplying 5 copies of a number, $x\mapsto x^5$.
What if you want to solve the equation $5^x=11$? You take a logarithm with base $5$, $x=\log_5(11)$. Again this is a solution by definition: $\log_5(11)$ is the unique real number $x$ such that $5^x=11$. To get a more intuitive feel for it, you can look at the curve $y=5^x$, notice that it is always increasing, and that it crosses $y=11$ somewhere between $1$ and $2$ (because $5^1=5$ and $5^2=25$). Ultimately the definition of $\log_5$ relies on the notion of inverting the more familiar operation of raising $5$ to a power, $x\mapsto 5^x$.
One way to get an intuitive feeling for the properties of logarithms is to see how the properties are derived from the more intuitive exponential function. For example, there is the familiar rule of exponents, $5^{a+b}=5^a\cdot 5^b$. This equation implies by the definition of the logarithm that $a+b=\log_5(5^a\cdot 5^b)$. On the other hand, $a=\log_5(5^a)$ and $b=\log_5(5^b)$ also by the definition of the logarithm, so the exponential identity becomes the logarithmic identity $\log_5(5^a)+\log_5(5^b)=\log_5(5^a\cdot 5^b)$. When $a$ and $b$ range over the real numbers, this implies the general product-to-sum identity, $\log_5(uv)=\log_5(u)+\log_5(v)$.
Everything comes from "switching $x$ and $y$," as Joe said. For example, every time $1$ is added to $x$, $y=5^x$ increases by a factor of five: $5^{x+1}=5\cdot 5^x$. Therefore, for the inverse, every time $x$ increases by a factor of $5$, $1$ is added to $y=\log_5(x)$.
Part of the answer to your question to when we would say "Let's take the $\log$" is whenever we are trying to solve for a quantity in an exponent, echoing Joe's answer somewhat (but devoid of the practical context given there). If you want to know when $(2t+1)^3 = 10$, a good first step is to say, "Let's take the cube root!" Analogously, if you want to know when $3^{2t+1}=10$, a good first step is to say, "Let's take the $\log$!"