Question

What is the solution set to this equation?
log4 (x + 3) + log4x =1

A. x=0
B. x=1
C. x=1 and x= -4
D. x=-3 and x=0

Answer 1

Answer:

B

Step-by-step explanation:

Using the rules of logarithms

log x + log y = log (xy)

$log_{b}$ x = n , then x = $b^{n}$

Given

$log_{4}$ (x + 3) + $log_{4}$ x = 1 , then

$log_{4}$ x(x + 3) = 1

x(x + 3) = $4^{1}$ = 4

x² + 3x = 4 ( subtract 4 from both sides )

x² + 3x - 4 = 0 ← in standard form

(x + 4)(x - 1) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 4 = 0 ⇒ x = - 4

x - 1 = 0 ⇒ x = 1

However, x > 0 , thus x = 1