Question

Log₂ (3 - x) + log₂ (2x) = 1​

Answer 1

The values of x in the logarithmic equation log₂ (3 - x) + log₂ (2x) = 1​ are x = 2.62 and x = 0.38

How to determine the value of x in the logarithmic equation

From the question, we have the following parameters that can be used in our computation:

log₂ (3 - x) + log₂ (2x) = 1​

Applying the product rule of logarithm, we have

log₂((3 - x) * 2x) = 1​

This gives

log₂(6x - 2x²) = 1​

So, we have

6x - 2x² = 2¹

Divide through by 2

So, we get

3x - x² = 1

This gives

x² - 3x + 1 = 0

Next, we solve for x using the quadratic formula

`x = (-b \pm √(b^2 - 4ac))/(2a)`

In x² - 3x + 1 = 0, we have a = 1, b = -3 and c = 1

So, we have

`x = (3 \pm √((-3)^2 - 4 * 1 * 1))/(2 * 1)`

`x = (3 \pm √(5))/(2)`

This gives

`x = (3 \pm 2.24)/(2)`

Expand

x = (3 + 2.24)/2 and x = (3 - 2.24)/2

So, we have

x = 2.62 and x = 0.38

Hence, the values of x in the logarithmic equation are x = 2.62 and x = 0.38