Question

Solve for x : log(3x) + log(x + 4) = log(15).Please explain it to me.​

Answer 1

Answer:

$x=1$

Step-by-step explanation:

Given:

$\log (3x)+\log(x+4)=\log (15)$

As the logs have no base, assume that the base is 10.

$\textsf{Apply log Product law}: \quad \log_ax + \log_ay=\log_axy$

$\implies \log_{10} (3x)+\log_{10}(x+4)=\log_{10} (15)$

$\implies \log_{10} \left(3x(x+4)\right)=\log_{10} (15)$

Expand the brackets:

$\implies \log_{10} \left(3x^2+12x\right)=\log_{10} (15)$

$\textsf{Apply the log Equality law}: \quad \textsf{if }\: \log_ax=\log_ay\:\textsf{ then }\:x=y$

$\implies 3x^2+12x=15$

Subtract 15 from both sides:

$\implies 3x^2+12x-15=0$

Factor out the common term 3:

$\implies 3(x^2+4x-5)=0$

Divide both sides by 3:

$\implies x^2+4x-5=0$

Split the middle term:

$\implies x^2+5x-x-5=0$

Factorize the first two terms and the last two terms separately:

$\implies x(x+5)-1(x+5)=0$

Factor out the common term (x + 5):

$\implies (x-1)(x+5)=0$

Therefore:

$x-1=0 \implies x=1$

$x+5=0 \implies x=-5$

As logs cannot be taken of negative numbers, $x=-5$ is an extraneous solution.  Therefore, the only valid solution is:  $x=1$

Answer 2

Answer:

x = 1

Explanation:

$\sf \rightarrow log(3x) + log(x + 4) = log(15)$

Rule: log(a) + log(b) = log(ab)

$\sf \rightarrow log(3x(x + 4)) = log(15)$

cancel out log on both sides

$\sf \rightarrow 3x(x + 4) = 15$

relocate constant variable

$\sf \rightarrow 3x^2 + 12x -15 = 0$

take 3 as a common factor

$\sf \rightarrow 3(x^2 + 4x -5) = 0$

divide both sides by 3

$\sf \rightarrow x^2 + 4x -5 = 0$

middle term split

$\sf \rightarrow x^2 + 5x -x-5 = 0$

factor common terms

$\sf \rightarrow x(x + 5) -1(x+5)= 0$

collect into groups

$\sf \rightarrow (x-1)(x+5)= 0$

set to zero

$\sf \rightarrow x-1 = 0 , \ x+5= 0$

relocate variables

$\sf \rightarrow x = 1, \ x = -5$

There must be a positive solution for log, so the solution is only x = 1