# What is the solution of equation log4 (x+1)+log3 (x+4)=1?

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To use the product property to the left side, we'll have to create matching bases;

log 4 (x+4) = log 3 (x+4)/log 3 4 => log 3 (x+4) = (log 3 4)*(log 4 (x+4))

The equation will become:

log4 (x+1)+(log 3 4)*(log 4 (x+4)) = log 4 4

(log 3 4)*(log 4 (x+4)) = log 4 4 - log4 (x+1)

(log 3 4)*(log 4 (x+4)) = log 4 [4/(x+1)]

(ln 4/ln 3)*(log 4 (x+4)) = log 4 [4/(x+1)]

(ln 4/ln3) `~=`

log 4 (x+4)^1.263 = log 4 [4/(x+1)]

Since the bases are matching, we'll use one to one property:

(x+4)^1.263 = [4/(x+1)]

(x+1)*(x+4)^1.263 = 4

The factors could be:

x+ 1 =1

x = 0 => (x+4)^1.263 = 4 impossible

x+1 = -1

x = -2 => (-2+4)^1.263 = -4 impossible

x + 1 = 2

x = 1 => (1+4)^1.263 = 2 impossible

x + 1 = -2

x = -3 => (-3+4)^1.263 = -2 impossible

x + 1 = 4 => (3+4)^1.263 = 1 impossible

x = 3

x + 1=-4

x = -5 => (-5+4)^1.263 = -1  => x = -5

We'll accept as solution of equation x = -5