What is the solution set to this equation?
log_4(x + 3) + log_4x = 1

x=1

Step-by-step explanation:

log_4(x + 3) + log_4x = 1

We know that loga(b) + loga(c) = loga(bc)

log_4(x + 3)x = 1

Raise each side to the base of 4

4^log_4(x + 3)x = 4^1

(x+3)x = 4

x^2 +3x = 4

Subtract 4 from each side

x^2 +3x -4 = 0

Factor

(x+4) (x-1) =0

Using the zero product property

x= -4  x=1

But x cannot be negative since logs cannot be negative

x=1

A..  x = 1.

Step-by-step explanation:

log_4(x + 3) + log_4x = 1

log_4 x(x + 3) = log_4 4

Removing the logs:

x(x + 3) = 4

x^2 + 3x - 4 = 0

(x + 4)(x - 1) = 0

x = 1, -4.

We can ignore the -4 as there is no log of a negative.

1d

2b

3 false

4 d

5 3200sq,ft

80+60+y=180

then divide the difference between 60 and y and add it to 180

so x =180 +( (y-60)/2)

x equals 200

step-by-step explanation:

8

d

step-by-step explanation: