Question

Log0, 5(x-2)-log0,5(10-x)=-1​

Answer 1

Ответ:

Объяснение:

Log(0.5,x-2)-log(0.5,10-x)=-1​

ODZ:

{x-2>0

{10-x>0

x∈(2;10)

$log_{0.5} }\frac{x-2}{10-x} =log_{0.5} 0.5^{-1}\\ \\ \frac{x-2}{10-x}=2\\ \\ x-2=20-2x\\ \\3x=22\\ \\ x=7\frac{1}{3}$

Answer 2

To solve the equation, let's first simplify the expression involving logarithms. Remember that the logarithm of zero is undefined, so we need to find the values of 'x' that make the expressions inside the logarithms greater than zero.

Given equation: log₀(5(x-2)) - log₀.₅(10-x) = -1

To combine the logarithms with the same base, we can use the logarithm identity:

logₐ(b) - logₐ(c) = logₐ(b/c)

Applying this identity, the equation becomes:

log₀.₅(5(x-2)/(10-x)) = -1

Now, we can rewrite the equation using the definition of logarithms. The equation logₐ(b) = c can be expressed as aᶜ = b.

So, we have:

₀.₅^(-1) = 5(x-2)/(10-x)

Now, solve for 'x':

2 = 5(x-2)/(10-x)

Next, let's clear the fraction by multiplying both sides by (10-x):

2(10-x) = 5(x-2)

Now, distribute on both sides:

20 - 2x = 5x - 10

Move all 'x' terms to one side:

5x + 2x = 20 + 10

Combine like terms:

7x = 30

Finally, solve for 'x':

x = 30/7

So, the solution to the equation is:

x ≈ 4.29