What is log2 (x+5) + log2 (x-1) =4?

Question

What is log2 (x+5) + log2 (x-1) =4?

Answer 1

log2 (x+5) + log2 (x-1) = 4
log2 ((x+5) (x-1) ) = log2 (2^4)
x^2 + 4 x - 5 = 16
x^2 + 4 x - 21 = 0
x=3 or x=-7
log is defined for positive numbers so x=3

Answer 2

Because you have a log2 as the base of the two expressions on the left, but not on the right, you can exponentiate both sides. Since the base of the log is 2, you can multiply both logs by 2, canceling them out. You would then need to change the value you are solving for to 2^4. The equation would go from log2(x+5) + log2(x-1)=4 to 2log2(x+5)+2log2(x-1)=2^4. The logs cancel leaving you with (x+5)(x-1)=16. Because addition between logs means multiplication you foil the expression.
x^2 + 4x - 5 = 16.
Set the expression equal to zero.
x^2 + 4x - 21 = 0
Factor.
(x + 7)(x - 3) = 0
x = -7 | x = 3
You can't take the log of a negative number, so your answer is only positive three. Hope this helps!

Answer 3

log2[(x + 5)(x - 1)] = 4
2^4 = (x + 5)(x - 1)
16 = x^2 - x + 5x - 5
0 = x^2 + 4x - 21
0 = (x + 7)(x - 3)

0 = x + 7
-7 = x

0 = x - 3
3 = x

x = -7 or 3

When you plug them back in, if you put in -7 you will be taking the log of a negative number which is impossible.

Your only answer is 3.

Answer 4

The sum of the logs is the log of their product
So
log_2(x+5)(x-1) = 4.
From the definition of logarithm,
(x+5)(x-1) = 16
since 2^4 = 16.
x²+4x-21=0.
(x+7)(x-3)= 0.
So x = -7 or x = 3.
But the first solution is invalid, because
it makes the argument of both the original
logarithms negative.
So the answer is x = 3.

Answer 5

log base 2 (x+5)(x-1) = 4
You mult logs.

(X+5)(X-1) = 2^4
x^2 + 4x - 5 = 16
x^2 + 4x - 21 = 0
(x + 7)(x - 3) = 0

x = 3 only

Answer 6

This may not be of help.

But I grieve for your poor soul...I hated logs so badly.