Log (base2)(x+3) + log(base2)(x-3) = 4?

Question

How do I solve for x?????

Answer 1

Log (base2)(x+3) + log(base2)(x-3) = 4
Log (base2)(x+3)(x-3) =4
Log (base2)(x^2-9)=4
(x^2-9)=4^2
x^2-9-16=0
x^2-25=0
x^2=25
x=+/-5

I hope this helps.:)

Answer 2

log a + log b = log ab for any base

so Log (base2) (x+3) (x-3)= 4

(x+3) (x-3)= x^2-9

so x^2-9 = 2^4 = 16

x^2 =25 x=+5 since x=-5 is impossible

check log (base2) (5+3) + log (base2 (5-3) =

log (base2) 8 + log (base 2) 2 = 3+1 =4


It works

Answer 3

log (base2) [x+3] + log (base2) [x-3] = 4

From the properties of logarithms,
log (baseA) [m] + log (baseA) [n] = log (baseA) [m][n]

Therefore:
log (base2) [x+3][x-3] = 4

Re-writing the equation in exponential form, we get:
2^4 = (x+3)(x-3)

Solving further,
16 = (x+3)(x-3)

(x+3)(x-3) can be reduced to (x^2 - 9) because it's the difference of two squares, the equation is now:
16 = x^2 - 9

Solving for x^2, we get:

x^2 = 16 + 9
x^2 = 25

Therefore, x = +/- 5

Answer 4

log (base 2)(x^2-9) = 4

2^4 = x^2-9

16 = x^2-9

25 = x^2

x = + or - 5

x = only 5 because -5 is impossible (you cannot take a negative logarithm.

Hope this helps you.
Demonsthenes

Answer 5

adding means multiplication so x+3 times x-3 equals 4 i believe