Answer:
The solutions are x = 1 and x = -5
Step-by-step explanation:
Hi there!
First, let´s write the equation:
log(3x) + log(x + 4) = log(15)
Apply logarithm property: log(3x) = log(3) + log(x)
log(3) + log(x) + log(x + 4) = log(15)
Substract log(3) from both sides of the equation
log(x) + log(x+4) = log(15) - log(3)
Apply logarithm property: log(15) - log(3) = log(15/3) = log(5)
log(x) + log(x + 4) = log(5)
Apply logarithm property: log(x) + log(x+4) = log(x (x+4)) = log(x² + 4x)
log(x² + 4x) = log(5)
Apply logarithm equality rule: if log(x² + 4x) = log(5), then x² + 4x = 5
x² + 4x = 5
Substract 5 from both sides
x² + 4x - 5 = 0
Using the quadratic formula (a = 1, b = 4, c = -5)
x = 1 and x = -5
The solutions are x = 1 and x = -5
Have a nice day!
Answer:
x=1
Step-by-step explanation:
The given equation is
[tex]\log(3x)+\log (x+4)=\log (15)[/tex]
We need to solve the equation for x.
Using the product property of logarithm we get
[tex]\log(3x(x+4))=\log (15)[/tex] [tex][\because \log (ab)=\log a+\log b][/tex]
[tex]\log(3x^2+12x)=\log (15)[/tex]
On comparing both sides we get
[tex]3x^2+12x=15[/tex]
[tex]3x^2+12x-15=0[/tex]
Find factors using grouping method.
[tex]3x^2+15x-3x-15=0[/tex]
[tex]3x(x+5)-3(x+5)=0[/tex]
[tex](x+5)(3x-3)=0[/tex]
Using zero product property we get
[tex]x+5=0\Rightarrow x=-5[/tex]
[tex]3x-3=0\Rightarrow x=1[/tex]
For x=-5, 3x=-15 and log is not defined for negative values.
It means x=-5 is an extraneous solution.
Therefore, x=1 is the only valid solution of the given equation.
