It will suffice to show that:
$$a<\log(127)$$
If we define:
$$f(x)=x^3-3x-3$$
A quick sign check shows us:
$$f(2)=2^3-3(2)-3=-1$$
$$f(3)=3^3-3(3)-3=15$$
So, we know $2<a<3$...and since $10^2<127<10^3$, we also know $2<\log(127)<3$. And so, we may begin the continued fraction representation of each by stating:
$$a\approx[2;\cdots]$$
$$\log(127)\approx[2;\cdots]$$
For the purpose of continuing the computation of the continued fraction representation of $\log(127)$, we will use:
$$\log(127)=2+\log(1.27)$$
To get the next partial quotient for $a$, we will make the substitution:
$$x=\frac{1}{x_1}+2$$
and we then find:
$$f_1\left(x_1\right)=x_1^3f\left(\frac{1}{x_1}+2\right)=x_1^3\left(\left(\frac{2x_1+1}{x_1}\right)^3-3\left(\frac{2x_1+1}{x_1}\right)-3\right)=-x_1^3+9x_1^2+6x_1+1$$
We then observe:
$$f_1(9)=55$$
$$f_1(10)=-39$$
And so we have:
$$a\approx[2;9,\cdots]$$
And using the iterative method outlined
>>here<<, we confirm that:
$$1.27^9<10<1.27^{10}$$
Hence:
$$\log(127)\approx[2;9,\cdots]$$
So, we find the next partial quotient for $a$:
$$f_2\left(x_2\right)=x_2^3f_1\left(\frac{9x_2+1}{x_2}\right)=x_2^3\left(-\left(\frac{9x_2+1}{x_2}\right)^3+9\left(\frac{9x_2+1}{x_2}\right)^2+6\left(\frac{9x_2+1}{x_2}\right)+1\right)=55x_2^3-75x_2^2-18x_2-1$$
We find:
$$f_2(1)=-39$$
$$f_2(2)=103$$
Thus:
$$a\approx[2;9,1,\cdots]$$
And we then confirm:
$$\left(\frac{10}{1.27^9}\right)^1<1.27<\left(\frac{10}{1.27^9}\right)^2$$
Hence:
$$\log(127)\approx[2;9,1,\cdots]$$
Continuing this process, we eventually find:
$$a\approx[2;9,1,1,1,2,1,\cdots]$$
$$\log(127)\approx[2;9,1,1,1,2,4,\cdots]$$
And so we may now conclude:
$$a<\log(127)$$
