Question

Log2 (x+2) + log2 (x-1) = 5

Answer 1

ОДЗ: x + 2 > 0  ⇒ x > - 2
         x - 1 > 0   ⇒ x > 1
Окончательно : x > 1
$log _{2} (x+2)+log _{2}(x-1)=5\\\\log _{2} (x+2)(x-1)=5\\\\(x+2)(x-1)=32\\\\ x^{2} -x+2x-2-32=0\\\\ x^{2} +x-34=0\\\\D=1 ^{2}-4*1*(-34)=1+136=137 \\\\ x_{1} = \frac{-1+ \sqrt{137} }{2}\\\\ x_{2}= \frac{-1- \sqrt{137} }{2}$
x₂ - не подходит
Ответ :
$x= \frac{-1+ \sqrt{137} }{2}$

Answer 2

To solve this equation, we can use the properties of logarithms to combine the two logarithmic terms into a single logarithm:

log2[(x+2)(x-1)] = 5

We can then rewrite this equation in exponential form:

2^5 = (x+2)(x-1)

32 = x^2 + x - 2

We can then rearrange this equation into standard quadratic form:

x^2 + x - 34 = 0

We can then use the quadratic formula to solve for x:

x = (-1 ± sqrt(1 + 4*34))/2

x = (-1 ± sqrt(137))/2

Therefore, the solutions to the equation are:

x ≈ 4.39 or x ≈ -5.39

Note that we need to check whether these solutions satisfy the original equation. We can do this by plugging them into the equation and verifying that both sides are equal. We find that only x ≈ 4.39 satisfies the equation, so this is the only valid solution.