Answer: [tex]x=\frac{31}{2}[/tex]
Step-by-step explanation:
[tex]2log_a(x+2)=log_a(x+9)+log_a(x-3)[/tex]
The first thing we are going to do is to get rid of the 2 in front of the first logarithm. To do this, we have the power property that says:
[tex]log_aX^n=nlog_aX[/tex]
Basically, the number in front, is the power of the number in the middle.
Let's rewrite this.
[tex]log_a(x+2)^2=log_a(x+9)+log_a(x-3)[/tex]
Now, in order to add logarithms with the same base (in this case "a"), we write the same base of the logarithm, and multiply their values.
[tex]log_a(x+2)^2=log_a(x+9)(x-3)[/tex]
Multiply the parentheses.
[tex]log_a(x+2)^2=log_a(x^2-3x+9x-27)[/tex]
Combine like terms;
[tex]log_a(x+2)^2=log_a(x^2+6x-27)[/tex]
Solve the binomial on the left side.
[tex]log_a(x^2+4x+4)=log_a(x^2+6x-27)[/tex]
When we have logarithms with same base, we equal their values.
[tex]x^2+4x+4=x^2+6x-27[/tex]
Subtract [tex]x^2[/tex]
[tex]x^2-x^2+4x+4=x^2-x^2+6x-27\\4x+4=6x-27[/tex]
Subtract 6x
[tex]4x-6x+4=6x-6x-27\\-2x+4=-27[/tex]
Subtract 4
[tex]-2x+4-4=-27-4\\-2x=-31[/tex]
Divide by -2.
[tex]x=\frac{-31}{-2}\\ x=\frac{31}{2}[/tex]
