Just hammer on the equation for a bit
2^x = 10x
(-10\ln2)*(-ln2*x) * e^(-ln2*x) = 1
(-ln2*x) * e^(-ln2*x) = -(ln2)/10
Now apply W
-ln2*x = W(-(ln2)/10)
x = -W(-(ln2)/10)/ln2
-1/e < -(ln2)/10 < 0 is true, so then the principal branch W_0 and branch W_-1 will both be real numbers.
Then Wolfram Alpha evaluates x as solutions x_0 ~= 0.107755 and x_-1 ~= 5.87701