...Ok the first one is: 7^2x+3 = 5^3x-2

I Understand that I can put them in log form, (2x+3)log 7 = (5x-2)log5

but, my teacher mentioned we should use the natural log? So how do I go about doing so?

doesn't matter ... same a log base 10.

(2x+3)ln 7 = (3x-2)ln 5

solve for x

2. 8^x2 * 4^-3x = 1/2

How I did it:

2^(3x^2) * 2^-6x = 1/2

2^(3x^2-6x) = 2^(-1)

3x^2 - 6x = -1

3. 8-2|6-2x| ≤ -4

-8 -8

-2|6-2x| ≤ -12

divide both sides by -2 ... dividing by a negative value changes the direction of the inequality.

|6-2x| > 6

4. Last one:

2^x+1 * 8^-x = 4

2^(x+1) * 2^(-3x) = 2^2

2^(1-2x) = 2^2

exponents are equal ...