Let's find out. We'll use the good 'ol Rule of 72.

According to the compound interest formula, if a principal amount of money $P$ is put in an account at an interest rate $r$ that is compounded continuously, then the amount in the account at time $t$ is

\begin{equation}\label{eq_cont_comp} A = Pe^{rt} \end{equation} Furthermore, the time it takes to double your principal is \begin{equation} \label{eq_t1} t_1 = \dfrac{\ln{2}}{r} \end{equation} Equation \eqref{eq_t1} follows from \eqref{eq_cont_comp} as follows: \begin{equation*} \begin{array}{rcl} 2P & = & Pe^{rt_{1}} \\ \ln{2} & = & rt_{1} \\ t_{1} & = & \dfrac{\ln{2}}{r} \\ \end{array} \end{equation*}

One might have thought that if $t_1$ were the time it took to double one's principal the first time, then $n t_1$ would be the time it took to double that principal $n$ times, but as we shall see, this is not so. If it were, then the number of doublings would be a linear function of $n.$ Instead, we shall see that $n t_1$ is the time it takes to double your principal $2^{n-1}$ times, an exponential function of $n.$ Let's state this principle formally:

If $t_1$ is the time it takes to double some principal one time under compound interest, then $nt_1$ is the time it takes to double the principal $2^{n-1}$ times.

It follows that the sequence $\{t_n\}$ given by \begin{equation} \label{eq_tn} t_n = nt_1, \end{equation} the first $n$ multiples of $t_1,$ are the milestones at which the amount in the account will be $A=2^nP,$ a $\left(2^n-1\right)\times100\%$ increase of $P.$

Equation \eqref{eq_tn} follows from \eqref{eq_cont_comp} as follows: \begin{equation*} \begin{array}{rcl} 2^{n-1}\left(2P\right) &=& Pe^{rt_n} \\ 2^nP &=& Pe^{rt_n} \\ \ln{2^n} &=& rt_n \\ t_n &=& \dfrac{\ln{2^n}}{r} \\ &=& \dfrac{n\ln{2}}{r} \\ &=& nt_1 \end{array} \end{equation*}

Equation \eqref{eq_tn} is a subtle result. For example, it states that if it takes so long to double one's money, a $100\%$ increase of one's principal, then in six times that period, one will not double one's principal six times, an $1100\%$ increase, but $32$ times, a $6300\%$ increase. This represents a doubling of the principal 32 times, or an amount of $64P.$ In five times that period one will not experience a "mere" $900\%$ increase in principal, a doubling of your principal $5$ times, but rather, a $3100\%$ increase, a doubling of your principal $16$ times.

Consequently, reducing $t_1$ improves the situation dramatically for any $t_n.$ By equation \eqref{eq_t1}, this reduction can only be done by increasing $r.$ In particular, $t_1$ is independent of $P$ by equation \eqref{eq_t1}. For example, if two investments of different size are made at the same rate of return, then they will both take equally long to double.

We can find the rate at which the amount in the account is increasing at time $t$ by differentiating equation \eqref{eq_cont_comp}: \begin{equation} m(t) = \dfrac{dA}{dt} = Pre^{rt} \end{equation} At integral multiples of $t_1,$ this rate becomes \begin{equation} \label{eq_da_dt_at_tn} m_n = \left.\dfrac{dA}{dt}\right\vert_{t = n t_1} = 2^n P r \end{equation}

As these rates are in units of $\frac{\mathrm{money}}{\mathrm{time}},$ e.g. $\frac{\$}{\mathrm{year}},$ they can be interpreted as the "price" of time. We can now figure out what your time is worth.

Suppose your principal is $P=\$1$ and the annual interest rate is $\ln{2} \approx.69.$ Then by equation \eqref{eq_t1} the time it takes to double $P$ is $1$ year $(= t_1).$ At that time, a year costs $m_1 = 2\ln{2}\approx\frac{$1.39}{\textrm{yr}}$ by equation \eqref{eq_da_dt_at_tn}. In contrast, at the end of year $6$ $(= t_6 = 6t_1),$ a year costs $m_6 = 2^6 \ln{2} =64 \ln{2} \approx \frac{$44.36}{\textrm{yr}}.$

That's worth repeating. The cost of a year went up $3100\%$ in 5 years, from $\$1.39$ to $\$44.36.$

So what is your time worth? A lot. And its price grows exponentially with time given by equation \eqref{eq_da_dt_at_tn}.