Answer:
A) [tex]x=-6[/tex]
Step-by-step explanation:
[tex]2\log_5 (x+1)=2[/tex]
We can use logarithm properties to write it as:
[tex]\log_5 (x+1)^2=2[/tex]
On removing [tex]log[/tex] it gives:
[tex](x+1)^2=5^2[/tex]
Taking square root both sides.
[tex]\sqrt{(x+1)^2}=\sqrt{5^2}[/tex]
[tex](x+1)=\pm 5[/tex]
Solving for [tex]x[/tex] we get:
[tex]x+1=5[/tex] and [tex]x+1=-5[/tex]
Subtracting by 1 to both sides of both the above equations.
[tex]x+1-1=5-1[/tex] and [tex]x+1-1=-5-1[/tex]
[tex]x=4[/tex] and [tex]x=-6[/tex]
So we have got two solutions for the logarithmic equation.
Now we check the validity of the solutions by plugging values in the equation [tex]2\log_5 (x+1)=2[/tex]
Plugging [tex]x=4[/tex]
[tex]2\log_5 (4+1)=2[/tex]
[tex]2\log_5 5=2[/tex]
[tex] 2(1)=2[/tex] [As [tex]log_5 5=1[/tex]]
[tex] 2=2[/tex]
So, thats a valid solution.
Plugging [tex]x=-6[/tex]
[tex]2\log_5 (-6+1)=2[/tex]
[tex]2\log_5 (-5)=2[/tex]
As [tex]\log_5 (-5)[/tex] is undefined, so [tex]x=-6[/tex] is not a valid solution which means its extraneous.
