why when we change the cost of every edge in G as c'= log17（c）,every MST in G is still an MST in G′ (and vice versa)?

Well, its pretty easy to understand...let's see if I can break it down for you:

c` = log_17(c) // here 17 is base

log may not be linear function...but we can say that:

log_b(x) > log_b(y) if x > y and b > 1 (and of course x > 0 and y > 0)

I hope you get the equation I've written...In words in means, consider a base "b" such that b > 1, then log_b(x) would be greater than log_b(y) if x > y.

So, if we apply this rule in your costs of MST of G, then we see that the edges those were selected for G, would still produce the least possible edges to construct MST G' if c' = log_17(c) // here 17 is base.

UPDATE: As I can see you've problem understanding the proof, I'm elaborating a bit:

I guess, you know MST construction is greedy. We're going to use kruskal's algo to proof why it is correct.(In case, you don't know, how kruskal's algo works, you can read it somewhere, or just google it, you'll find millions of resources). Now, Let me write some steps of kruskal's edge selection for MST of G:

// the following edges are sorted by cost..i.e. c_0 <= c_1 <= c_2 ....

c_0: A, F // here, edge c_0 connects A, F, we've to take the edge in MST
c_1: A, B // it is also taken to construct MST
c_2: B, R // it is also taken to construct MST
c_3: A, R // we won't take it to construct to MST, cause (A, R) already connected through A -> B -> R
c_4: F, X // it is also taken to construct MST
...
...
so on...

Now, when constructing MST of G', we've to select edges which are in the form c' = log_17(c) // where 17 is base

Now, if we convert the edges using log of base 17, then c_0 becomes c_0', c_1 becomes c_1' and so on...

But we, know that:

log_b(x) > log_b(y) if x > y and b > 1 (and of course x > 0 and y > 0)

So, we may say that,

log_17(c_0) <= log_17(c_1), cause c_0 <= c_1
in general, 
log_17(c_i) <= log_17(c_j), where i <= j

And now, we may say:

c_0` <= c_1` <= c_2` <= c_3` <= ....

So, the edge selection process to construct MST of G' would be:

// the following edges are sorted by cost..i.e. c_0` <= c_1` <= c_2` ....

c_0`: A, F // here, edge c_0` connects A, F, we've to take the edge in MST
c_1`: A, B // it is also taken to construct MST
c_2`: B, R // it is also taken to construct MST
c_3`: A, R // we won't take it to construct to MST, cause (A, R) already connected through A -> B -> R
c_4`: F, X // it is also taken to construct MST
...
...
so on...

Which is same as MST of G...

That proves the theorem ultimately....

I hope you get it...if not ask me in the comment what is not clear to you...

One way to characterize when a spanning tree T is a minimum spanning tree is that, for every edge e not in T, the cycle formed by e and edges of T (the fundamental cycle of e with respect to T) has no edge more expensive than e. Using this characterization, I hope you see how to prove that transforming the costs with any increasing function preserves minimum spanning trees.

There's a one line proof that this condition is necessary. If the fundamental cycle contained a more expensive edge, we could replace it with e and get a spanning tree that costs less than T.

It's less obvious that this condition is sufficient, since at first glance it looks like we're trying to prove global optimality from a local optimality condition. To prove this statement, let T be a spanning tree that satisfies the condition, let T' be a minimum spanning tree, and let G' be the graph whose edges are the union of the edges of T and T'. Run Kruskal's algorithm on G', breaking ties by favoring edges in T over edges not in T. Let T'' be the resulting minimum spanning tree in G'. Since T' is a spanning tree in G', the cost of T'' is not greater than T', hence T'' is a minimum spanning tree in G as well as G'.

Suppose to the contrary that T'' ≠ T. Then there exists an edge in T but not in T''. Let e be the first such edge considered by Kruskal's algorithm. At the time that e was considered, it formed a cycle C in the edges that had been selected from T''. Since T is acyclic, C \ T is nonempty. By the tie breaking criterion, we know that every edge in C \ T costs less than e. Observing that some edge e' in C \ T must have one endpoint in each of the two connected components of T \ {e}, we infer that the fundamental cycle of e' with respect to T contains e, which violates the local optimality condition. In conclusion, T = T'', hence is a minimum spanning tree in G.

If you want a deeper dive, this logic gets abstracted out in the theory of matroids.