Answer:
[tex]x=1[/tex]
Step-by-step explanation:
Given:
[tex]\log (3x)+\log(x+4)=\log (15)[/tex]
As the logs have no base, assume that the base is 10.
[tex]\textsf{Apply log Product law}: \quad \log_ax + \log_ay=\log_axy[/tex]
[tex]\implies \log_{10} (3x)+\log_{10}(x+4)=\log_{10} (15)[/tex]
[tex]\implies \log_{10} \left(3x(x+4)\right)=\log_{10} (15)[/tex]
Expand the brackets:
[tex]\implies \log_{10} \left(3x^2+12x\right)=\log_{10} (15)[/tex]
[tex]\textsf{Apply the log Equality law}: \quad \textsf{if }\: \log_ax=\log_ay\:\textsf{ then }\:x=y[/tex]
[tex]\implies 3x^2+12x=15[/tex]
Subtract 15 from both sides:
[tex]\implies 3x^2+12x-15=0[/tex]
Factor out the common term 3:
[tex]\implies 3(x^2+4x-5)=0[/tex]
Divide both sides by 3:
[tex]\implies x^2+4x-5=0[/tex]
Split the middle term:
[tex]\implies x^2+5x-x-5=0[/tex]
Factorize the first two terms and the last two terms separately:
[tex]\implies x(x+5)-1(x+5)=0[/tex]
Factor out the common term (x + 5):
[tex]\implies (x-1)(x+5)=0[/tex]
Therefore:
[tex]x-1=0 \implies x=1[/tex]
[tex]x+5=0 \implies x=-5[/tex]
As logs cannot be taken of negative numbers, [tex]x=-5[/tex] is an extraneous solution. Therefore, the only valid solution is: [tex]x=1[/tex]
