I am noob in convex optimization and I have some questions 
The expression 
seems nonconvex, where a_n and b_n are variables, c is a constant. is it possible to transform it into a convex expression??
(We know that 
can be convex since it can be transformed to 
.)
Anyone can give me some suggestions? thanks !!!
when you transform it to an<=c*log(1+bn) ,in order to be accepted by cvx,the left side must be convex
and the right must be concave,log(1+bn) are concave when (1+bn) are concave and >0; which have menthioned in Convex optimization in part3,不说了，急着出门。 
In your previous reply, if the a_n is a binary variable, is it possible to use a slack variable to replace d_n*b_n ?
Section 2.8 of https://www.fico.com/en/resource-access/download/3217 shows how to linearize the product of binary variable with continuous variable.
ok, i saw this pdf in your previous answer in other posts, maybe it can help, thanks.
You can do variable substitution. Firstly, you should define x_n =\frac{a_n}{\log(1+b_n)}, \frac{a_n}{\log(1+b_{n}^{max})} \leq x_n \leq \frac{a_n}{\log(1+b_{n}^{min})}. Secondly,
you can replace \frac{a_n}{\log(1+b_n)} in the expression with x_n. Then, you have a new expression \sum_{n=1}^{N}x_n \leq c, \\
\textbf{s.t.} \frac{a_n}{\log(1+b_{n}^{max})} \leq x_n \leq \frac{a_n}{\log(1+b_{n}^{min})}. Finally, the above expression is a convex set.
@Yenfy I presume you are assuming a_n \ge 0.
Perhaps your formulation assumes a_n and b_n do not appear elsewhere in the model?
How are a_n and b_n recovered from x_n?
I,m sorry, it’s my fault. I thought a_n was a parameter. If a_n is a variable, my above solution is wrong.
@Yenfy If a_n were a parameter and b_n is constrained to be >= 0, the easiest thing to do in CVX would be
sum(a.*inv_pos(log(1+b))) <= c
When I use inv_pos() in CVX (not this problem.), the result usually become infeasible. 
,but when I comment out the constraint contains inv_pos() expression, it can be solved and the result can satify the commented consraints. 
， that’s very confusing.
@TomK Please provide a reproducible example, clearly showing what you say.
y can be a vector.
help quad_over_lin
quad_over_lin Sum of squares over linear.
Z=quad_over_lin(X,Y), where X is a vector and Y is a scalar, is equal to
SUM(ABS(X).^2)./Y if Y is positive, and +Inf otherwise. Y must be real.If X is a matrix, quad_over_lin(X,Y) is a row vector containing the values of quad_over_lin applied to each column. If X is an N-D array, the operation is applied to the first non-singleton dimension of X. quad_over_lin(X,Y,DIM) takes the sum along the dimension DIM of X. A special value of DIM == 0 is accepted here, which is automatically replaced with DIM == NDIMS(X) + 1. This has the effect of eliminating the sum; thus quad_over_lin( X, Y, NDIMS(X) + 1 ) = ABS( X ).^2 ./ Y. In all cases, Y must be compatible in the same sense as ./ with the squared sum; that is, Y must be a scalar or the same size as SUM(ABS(X).^2,DIM). Disciplined convex programming information: quad_over_lin is convex, nonmontonic in X, and nonincreasing in Y. Thus when used with CVX expressions, X must be convex (or affine) and Y must be concave (or affine).
I want to minimize 
This is my code:
%%%%%%%%%%%%%%%%%%%%%%%%%%
cvx_begin
% cvx_solver SeDuMi
cvx_solver SDPT3
cvx_precision low
variable a(5)
variable V_slack(5)
minimize( sum( quad_over_lin(a,V_slack) ) )
…
The error becomes
