log2 (x+4) +log2 (x-3) =3

Question

Could someone walk me through the steps? I tried to factor to find x but it didn't work.

Answer 1

log₂ (x + 4) + log₂ (x – 3) = 3. Could someone walk me through the steps? I tried to factor to find x but it didn't work.

Detailed Solution:

log₂ (x + 4) + log₂ (x – 3) = 3 <== Apply: log A + log B = log AB where A = (x + 4) and B = (x – 3)

Therefore: log₂ (x + 4) + log₂ (x – 3) = 3 is the same as log₂ (x + 4) (x – 3) = 3

log₂ AB = 3 is the same as: AB = 2³

Therefore: log₂ (x + 4) (x – 3) = 3 is the same as (x + 4) (x – 3) = 2³

(x + 4) (x – 3) = 2³

x² – 3x + 4x – 12 = 8

x² + x – 12 – 8 = 0

x² + x – 20 = 0

(x – 4) (x + 5) = 0

x – 4 = 0 ==> x = 4 <== Valid Answer

x + 5 = 0 ==> x = –5 <== Invalid Answer

Check your answers in the original equation: log₂ (x + 4) + log₂ (x – 3) = 3

When x = 4: log₂ (x + 4) + log₂ (x – 3) = 3 ==> log₂ (4 + 4) + log₂ (4 – 3) = 3 ==> log₂ 8 + log₂ 1 = 3 ==> log₂ 2³ + log₂ 1 = 3 ==> 3 log₂ 2 + log₂ 1 = 3 ==> 3(1) + 0 = 3 ==> 3 = 3 <== Valid Answer

When x = –5: log₂ (x + 4) + log₂ (x – 3) = 3 ==> log₂ (–5 + 4) + log₂ (–5 – 3) = 3 ==> log₂ (–1) + log₂ (–8) = 3 ==> Invalid + Invalid = 3 ==> Invalid = 3 <== Invalid Answer

Answer 2

If I understand the problem correctly, it is:

log₂(x + 4) + log₂(x - 3) = 3

An important rule to remember for logarithms is:

log_b(M) + log_b(N) = log_b(M*N)

For this problem, let's say x + 4 = M, x - 3 = N, and 2 = b. So

log₂(x + 4) + log₂(x - 3) = log₂((x + 4)*(x - 3))

This means:

log₂((x + 4)*(x - 3)) = 3

Simplifying

log₂(x² + x - 12) = 3

Now we do

x² + x - 12 = 2³

x² + x - 12 = 8

x² + x - 20 = 0

(x + 5)(x - 4) = 0

x = -5, 4

Edit: As the commenter below me mentioned, plugging -5 in to the original equation results in taking logs of negative numbers (which you cannot do) so the only solution is 4.

If you need any more help, don't hesitate to ask!