Question

If you're good at logarithms please help me with question h and show full working out ty ;)

Answer 1

Answer: $x=\frac{31}{2}$

Step-by-step explanation:

$2log_a(x+2)=log_a(x+9)+log_a(x-3)$

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:

$log_aX^n=nlog_aX$

Basically, the number in front, is the power of the number in the middle.

Let's rewrite this.

$log_a(x+2)^2=log_a(x+9)+log_a(x-3)$

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.

$log_a(x+2)^2=log_a(x+9)(x-3)$

Multiply the parentheses.

$log_a(x+2)^2=log_a(x^2-3x+9x-27)$

Combine like terms;

$log_a(x+2)^2=log_a(x^2+6x-27)$

Solve the binomial on the left side.

$log_a(x^2+4x+4)=log_a(x^2+6x-27)$

When we have logarithms with same base, we equal their values.

$x^2+4x+4=x^2+6x-27$

Subtract $x^2$

$x^2-x^2+4x+4=x^2-x^2+6x-27\\4x+4=6x-27$

Subtract 6x

$4x-6x+4=6x-6x-27\\-2x+4=-27$

Subtract 4

$-2x+4-4=-27-4\\-2x=-31$

Divide by -2.

$x=\frac{-31}{-2}\\ x=\frac{31}{2}$