Question

Solve for x : log(3x) + log(x + 4) = log(15).

Please explain it to me.​

Answer 1

`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

x = 1

Step-by-step 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