Non riesco a fare questa equazione con logaritmi. Le basi sono diverse.
LOG(√(x + 2)) = 3 - LOG(2,x + 2)
cambio base
LN(x + 2)/2 = 3 - LN(x + 2)/LN(2)
LN(x + 2)/2 + LN(x + 2)/LN(2) = 3
LN(x + 2)·(1/LN(2) + 1/2) = 3
LN(x + 2) = 3/(1/LN(2) + 1/2)
LN(x + 2) = 6·LN(2)/(LN(2) + 2)
x + 2 = e^(6·LN(2)/(LN(2) + 2))
x = 2^(6/(LN(2) + 2)) - 2
Dal momento che il radicale ha indice due mi sembra logico ridurre a base due.
* - log(10, √(x + 2)) = 3 - log(2, x + 2) ≡
≡ log(2, x + 2) - log(2, (x + 2)^(1/2))/log(2, 10) = 3 ≡
≡ (log(2, 10))*log(2, x + 2) - log(2, x + 2)/2 = 3*log(2, 10) ≡
≡ (log(2, 10) - 1/2)*log(2, x + 2) = 3*log(2, 10) ≡
≡ log(2, x + 2) = 3*log(2, 10)/(log(2, 10) - 1/2) ≡
≡ 2^log(2, x + 2) = 2^(3*log(2, 10)/(log(2, 10) - 1/2)) ≡
≡ x + 2 = 10^(log(10, 64)/log(10, 50)) ≡
≡ x = 10^(log(10, 64)/log(10, 50)) - 2 ~= 9.56386
Vedi anche
http://www.wolframalpha.com/input?i=-log%2810.0%2C%E2%88%9A%28x--2%29%29%3D3.0-log%282.0%2Cx--2%29