Pre-cal Help?

Question

can someone please solve for x

log(3)x+log(3)(x+4)=log(3)4

Answer 1

log_3 x + log_3 (x+4) = log_3 4

First, combine the expressions on the left using the laws of logarithms:

log_3 (x²+4x) = log_3 4

Exponentiate both sides:

x²+4x=4

Add 4 to both sides:

x²+4x+4=8

Factor the left:

(x+2)² = 8

Take the square roots of both sides:

x+2 = ±2√2

Subtract 2:

x=-2±2√2

This gives two solutions. However, the negative solution cannot be correct, because if x is a negative number, log_3 x is not defined. Therefore, the sole remaining solution is:

x=-2+2√2

And we are done.

Answer 2

Adding logs is the same as multiplying the arguement. So the next line goes
log(3) x(x+4) = log (3) 4
Since the log base 3 of each side is equal, the sides themselves are equal, so you can drop the log 3 and equate the arguement:
x(x+4) = 4
This is a quadratic equation, put it in standard form
x^2 + 4x - 4 = 0
put in the quadratic formula
x = (-4 +/- sqrt (16 + 16))/2
x = (-4 +/- 4 sqrt 2) / 2
Discard the negative root, as logs can only be positive
x = (-4 + 4 sqrt 2)/2
x = -2 + 2 sqrt 2
x = 2(-1 + sqrt 2)
x = 2(sqrt 2 - 1)

Answer 3

Since all bases are the same,

x(x+4) = 4
x^2 + 4x - 4 = 0

x = -4 + - sq rt (16 - 4(1)(-4) / 2(1)

x = - 4 + - sq rt (16 + 16) / 2

x = -4 + - sq rt 32 / 2

x = -4 + - 4 sq rt 2 / 2

x = - 2 + 2 sq rt 2

No negative answers for logs.

Answer 4

log(3)x+log(3)(x+4)=log(3)4
log(3)(x²+4x)=log(3)4
x²+4x-4=o
x=0.828427124 or x=-4.828427125