Question

A) (1/(2-log2x)) + (1/(1+log2x)) = 3/2 b) (1/(1-log3x)) + (1/(1+log3x)) = 3/2

Answer 1

Ответ: во вложении Объяснение:

Answer 2

To solve the given equations, we need to find the value(s) of 'x' that satisfy the equations:

A) $\frac{1}{{2-\log_2 x}} + \frac{1}{{1+\log_2 x}} = \frac{3}{2}$

B) $\frac{1}{{1-\log_3 x}} + \frac{1}{{1+\log_3 x}} = \frac{3}{2}$

Let's solve each equation one by one:

A) $\frac{1}{{2-\log_2 x}} + \frac{1}{{1+\log_2 x}} = \frac{3}{2}$

To simplify this equation, let's find a common denominator:

$\frac{(1+\log_2 x) + (2-\log_2 x)}{{(2-\log_2 x)(1+\log_2 x)}} = \frac{3}{2}$

Combine like terms in the numerator:

$\frac{3}{{(2-\log_2 x)(1+\log_2 x)}} = \frac{3}{2}$

Now, cross-multiply:

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

$6 = 2 - (\log_2 x)^2$

Rearrange the equation:

$(\log_2 x)^2 = 2 - 6$

$(\log_2 x)^2 = -4$

Now, take the square root of both sides (note that log values must be positive):

$\log_2 x = \pm \sqrt{-4}$

The square root of a negative number is not a real number. Hence, there are no real solutions for equation A).

B) $\frac{1}{{1-\log_3 x}} + \frac{1}{{1+\log_3 x}} = \frac{3}{2}$

To simplify this equation, let's find a common denominator:

$\frac{(1+\log_3 x) + (1-\log_3 x)}{{(1-\log_3 x)(1+\log_3 x)}} = \frac{3}{2}$

Combine like terms in the numerator:

$\frac{2}{{(1-\log_3 x)(1+\log_3 x)}} = \frac{3}{2}$

Now, cross-multiply:

$2 \times 2 = (1-\log_3 x)(1+\log_3 x)$

$4 = 1 - (\log_3 x)^2$

Rearrange the equation: