Is it not simply a question of saying that 2^10 = 1024 is close to 10^3 - 1000. Take 10th roots of both sides and 10^(3/10) = 1.995262... is close to 2.
And therefore we have 1 = 10^0, 2 ≈ 10^0.3, 4 ≈ 10^0.6, 8 ≈ 10^0.9, 16 ≈ 10^1.2, 32 ≈ 10^1.5, 64 ≈ 10^1.8, 128 ≈ 10^2.1, 256 ≈ 10^2.4, and 512 ≈ 10^2.7.
Divide by 10s and we get 1 = 10^0, 2 ≈ 10^0.3, 4 ≈ 10^0.6, 8 ≈ 10^0.9, 1.6 ≈ 10^0.2, 3.2 ≈ 10^0.5, 6.4 ≈ 10^0.8, 1.28 ≈ 10^0.1, 2.56 ≈ 10^0.4, and 5.12 ≈ 10^0.7.
Reorder and we've got the initially surprising 1 = 10^0, 1.28 ≈ 10^0.1, 1.6 ≈ 10^0.2, 2 ≈ 10^0.3, 2.56 ≈ 10^0.4, 3.2 ≈ 10^0.5, 4 ≈ 10^0.6, 5.12 ≈ 10^0.7, 6.4 ≈ 10^0.8 and 8 ≈ 10^0.9.
Then work the 5^10 and 7^13 examples to show that it happens whenever a^b is close to a power of 10.
Leave out the other base observation.
What's more tricky, I find, is communicating to people what I'm trying to achieve. I found the same problem with the "photocopy a mirror" problem, and the "Four Points, Two Distances" problem. It's not a question of doing the sums, it's a problem of making the structure of the solution accessible.
But in this case I think I'm getting there.