Question

PLEASE HELP ME QUICK

I DONT WANT TO FAIL !!

Answer 1

Answer: x ≥ 2

Explanation:

Since both of these logarithms have the same base, we can use the one-to-one property. Then we will solve as normal.

One-to-one-property:

3 - x ≤ x - 1

Add x to both sides of the equation:

3 ≤ 2x - 1

Add 1 to both sides of the equation:

4 ≤ 2x

Divide both sides of the equation by 2:

2 ≤ x

Flip to rewrite:

x ≥ 2

Please note that the log of a negative number is undefined. For example, if we input 5 (which is greater than two) 3 - 5 = -2 and
`log_6(-2)` is undefined.

Answer 2

`x\ge 2`

Explanation:

For this type of problem, we can take "6" and raise it to the power of both sides.

`6^{\text{log}_6(3-x)}\le6^{\text{log}_6(x-1)}`

which simplifies to:

`3-x\le x-1`

which makes sense since generally speaking, the logarithmic function is increasing (if the base is greater than one), so the inside must be less. Now from here we can add x to both sides and add 1 to both sides:

`4\le2x`

Now just divide both sides by 2:

`2\le x`

Which can also be written as:

`x\ge 2`

One important thing to note is this inequality is technically incomplete as the (3-x) must remain greater than zero and (3-x) cannot equal 1 nor can (x-1) equal 1 as when any of those conditions are true, one of the logarithms cannot be defined so we can't really compare them.