Solve the pair of simultaneous equation. log(y_x)=0 and 2logy=log(21+x)?

Given equations are:
1. log(y) = x
2. 2log(y) = log(21 + x)

From equation 1, we can rewrite it as:
y = 10^x

Substitute y = 10^x into equation 2:
2log(10^x) = log(21 + x)
2x = log(21 + x)
10^(2x) = 21 + x

Now, we can solve for x by substituting y = 10^x back into the equation:
10^(2x) = 21 + 10^x
Let 10^x = t
t^2 = 21 + t
t^2 - t - 21 = 0
(t - 4)(t + 5) = 0
t = 4 or t = -5

Since t = 10^x, we can solve for x:
10^x = 4 or 10^x = -5 (not possible as x is a real number)

Therefore, x = log(4) = 0.602

Now, substitute x back into equation 1 to find y:
y = 10^0.602
y ≈ 4.0

Therefore, the solution to the pair of simultaneous equations is x = 0.602 and y ≈ 4.