Question

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

Answer 1

Answer:
`x=(31)/(2)`

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=(-31)/(-2)\\ x=(31)/(2)`