how to solve log2 (x-4)-log2 (2/x) =5

Question

Log₂ (x-4) - log₂ (2/x) = 5

Answer 1

log₂(x-4) - log₂(2/x) = 5

 

log₂[(x-4) ÷ (2/x)] = 5

 

log₂[x(x-4)/2] = 5

 

x(x-4)/2 = 2⁵

 

x(x-4) = 64

 

x² - 4x - 64 = 0

 

By the quadratic formula, x = (4 ± √272)/2

 

                                          = 2 ± 2√17

 

                                          ≈ 10.2, -6.2

 

-6.2 is extraneous since we can't take the logarithm of a negative number.

 

So, the only solution is 2 + 2√17.

 

 

Answer 2

Hi Patricia, we can solve this using the rules for logs. we can break the problem into separate steps.

 

Step 1: 

 

Rewrite log₂ (2/x) using division rule, log_b (x/y) = log_b (x) - log_b (y)

 

log₂ (2/x) = log₂ (2) - log₂ (x)

 

log₂ (2) = 1, then

 

log₂ (2/x) = 1 - log₂ (x)

 

Step 2:

 

Replace log₂ (2/x) in the equation:

 

log₂ (x - 4) - (1 - log₂ (x) = 5

 

log₂ (x - 4) - 1 + log₂ (x) = 5

 

Step 3:

 

Add 1 to both sides of the equation to get the logs by themselves:

 

log₂ (x - 4) + log₂ (x) = 6

 

Step 4:

 

Rewrite log₂ (x - 4) + log₂ (x) using reverse log multiplication rule. 

 

Log multiplication rule: log_b (x * y) = log_b (x) + log_b (y)

 

This gives:

 

log₂ (x * (x - 4)) = 6

 

Step 5:

 

Use the inverse function of the exponential function.

 

If y = log_b (x), then x = b^y

 

log₂ (x * (x - 4)) = 6

 

The inverse function gives:

 

x * (x - 4) = 2⁶

 

Step 6:

 

Solve for x.

 

x² - 4x = 64

 

x² - 4x - 64 = 

 

Use the quadratic equation to solve, where a = 1, b = -4 and c = -64.

 

x = (-(-4) ± √((-4)² - 4 * (1) * (-64))) / 2 * (1)

 

x = (4 ± √( 16 + 256)) / 2

 

x = (4 ± √( 16 + 16 * 16)) / 2

 

x = (4 ± √( 16 * (1 + 16))) / 2

 

x = (4 ± 4 * √17) / 2

 

x = 2 ± 2 * √17

 

This gives two solutions, x = 2 + 2 √17 and x = 2 - 2√17

 

x = 10.24621124 and x = -6.246211251

 

Solution:

 

x = 10.24621124

 

We can discard x = -6.246211251 because logarithms cannot be negative. 

 

Questions?