Question

Iog2 (x-2)+log2 (x-3)=1

Answer 1

(x-2) (x-3)=2
x^2-3x-2x+6=2
x^2-5x+4=0
D=(-5)^2-16=25-16=9
x1=4
x2=1

Answer 2

Log2(x-2)(x-3)=log2 (2)
(x-2)(x-3)=2
x^2-3x-2x+6=2
x^2-5x+4=0
x1=4; x2=1(за т. Виета)

Answer 3

To solve the equation $\log_2(x-2) + \log_2(x-3) = 1$, we can use the properties of logarithms.

The sum of logarithms is the same as the logarithm of the product, so we can rewrite the equation as:

$\log_2[(x-2)(x-3)] = 1$

Next, we use the property of logarithms that states if $\log_a(b) = c$, then $a^c = b$.

So, in our case:

$2^1 = (x-2)(x-3)$

Simplifying:

$2 = (x-2)(x-3)$

Now, we have a quadratic equation. Let's expand it:

$2 = x^2 - 5x + 6$

To solve for x, we rearrange the equation to bring all terms to one side:

$x^2 - 5x + 4 = 0$

Now, we factor the quadratic:

$(x-4)(x-1) = 0$

Setting each factor to zero and solving for x:

$x-4 = 0$ or $x-1 = 0$

$x = 4$ or $x = 1$

However, we need to check these solutions in the original equation because the logarithm of a negative or zero value is not defined.

Checking $x = 4$:

$\log_2(4-2) + \log_2(4-3) = \log_2(2) + \log_2(1)$

Since $\log_2(1) = 0$, the equation is valid for $x = 4$.

Checking $x = 1$:

$\log_2(1-2) + \log_2(1-3) = \log_2(-1) + \log_2(-2)$

The logarithm of negative numbers is not defined in the real number system, so $x = 1$ is not a valid solution.

Thus, the only valid solution is $x = 4$.