The answer is 6.

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We start off by noticing that $4^{x}=(2^{x})^{2}$ and $5^{x}=\large\frac{10^{x}}{2^{x}}$ .

We now ask ourselves, "What digits could $2^{x}$ start with, so that $5^{x}$ starts with $1$ ?".

It's easy to see that the first, say, 4 digits are guaranteed to be within $[5000,9999]$ - if they're not, the first digit of the reciprocal will be at least $2$ . Note that we're not looking for a strictly equivalent condition, just a bound large enough to hold all the possibilities, but small enough to remain useful. This will hopefully come completely clear in a bit.

Now that we have our bound for $2^{x}$ 's first digits due to $5^{x}$ starting with $1$ , we find what $4^{x}$ gives us:

$\large10^{k}<(2^{x})^{2}<2\cdot10^{k}$

If $k$ is even: (with $a=\frac{k}{2}$ )

$\large10^{a}<2^{x}<\sqrt{2}\cdot10^{a}$ , thus $2^{x}$ must start with some 4 digits between the first 4 digits of $1$ and $\sqrt{2}$ , namely $[1000,1415]$ .

If $k$ is odd: (with $a=\frac{k-1}{2}$ )

$\large\sqrt{10}\cdot10^{a}<2^{x}<\sqrt{20}\cdot10^{a}$ , resulting in $2^{x}$ starting between $\sqrt{10}$ and $\sqrt{20}$ 's first digits, $[3162,4473]$ .

It's now clear it's impossible for $4^{x}$ and $5^{x}$ to both start with 1, since $2^{x}$ 's first 4 digits would have to be within both $[5000,9999]$ and $([1000,1415] \text{ or } [3162,4473])$ .

Doing the same procedure for the numbers $2$ through $9$ , only $2$ and $4$ have the two intervals join at any point - in the case of $2$ it's sufficient for $2^{x}$ to start between $[4473,4999]$ , and in the case of $4$ , the interval is $[2000,2236]$ .

Examples of such powers of $4$ and $5$ :

$4^{52}\approx2.028241\cdot10^{31}$

$5^{52}\approx2.220446\cdot10^{36}$

$4^{11}=4194304$

$5^{11}=48828125$

Thus the answer is $2+4=6$ .