How do I solve these inequalities:

e^3x>e^x-1

-6log (subscript of 4) x > -24

& how do I solve these:

8^x+1=3

e^x+2=50

8logx=16

I really don't understand how to solve these so can someone please explain them to me?

e^3x>e^x-1

you probably meant:
e^(3x) > e^(x-1), (huge difference from what you typed)
ln both sides
ln (e^(3x)) > ln e^(x-1)
3x > x-1
2x > -1
x > -1/2

PROOF:
http://www.wolframalpha.com/input/?i=plot+y+%3D+e%5E%283x%29+%2C+y+%3D+e%5E%28x-1%29+from+-1+to+0

-6log (subscript of 4) x > -24
divide both sides by -6
log4 x < 4
x < 4
but remember we can only take logs of positive numbers, so
0 < x < 4 , also remember that log(a number between 0 and 1 is negative, and multiplying it by -6 would make the result positive.

so 1 < x < 256

check: notice the graph of y = log4 x
is below y = 4 from 1 to 256

http://www.wolframalpha.com/input/?i=plot+y+%3D+4+%2C+y+%3D+log%28x%29%2Flog%284%29+from+.1+to+300

Now for the equations:
8^x+1=3
the way you typed it ...
8^x = 2
x = 1/3 , since the cuberoot of 8 is 2

the way you typed it:
e^x+2=50
e^x = 48
ln both sides
ln (e^x) = ln 48
x lne = ln 48
x = ln 48 , since lne = 1

8logx=16
divide both sides by 8
log x = 2
which means:
10^2 = x
x = 100

To solve these inequalities and equations, we will go step by step:

1. e^3x > e^x - 1 inequality:
- Start by subtracting e^x from both sides: e^3x - e^x > -1
- Now, factor out e^x: e^x (e^2x - 1) > -1
- As e^x is always positive, we can divide both sides by e^x without changing the direction of the inequality: e^2x - 1 > -1
- Add 1 to both sides: e^2x > 0
- Since e^2x is positive for all real values of x, this inequality holds true for any real number.

2. -6 log(sub4)x > -24 inequality:
- Divide both sides by -6 (note: dividing by a negative number reverses the inequality direction): log(sub4)x < 4
- Rewrite in exponential form: 4^(log(sub4)x) < 4^4
- Simplify: x < 256

Now, let's move on to solving the equations:

3. 8^(x+1) = 3 equation:
- Rewrite 8 as 2^3 and 3 as 2^log(sub2)(3): (2^3)^(x+1) = 2^log(sub2)(3)
- Apply the power rule of exponents: 2^(3(x+1)) = 2^log(sub2)(3)
- Since the bases are the same, the exponents must be equal: 3(x+1) = log(sub2)(3)
- Solve for x: x = (log(sub2)(3) - 3) / 3

4. e^(x+2) = 50 equation:
- Take the natural logarithm (ln) of both sides: ln(e^(x+2)) = ln(50)
- Apply the logarithm property: x + 2 = ln(50)
- Solve for x: x = ln(50) - 2

5. 8log(x) = 16 equation:
- Divide both sides by 8: log(x) = 2
- Rewrite in exponential form: x = 10^2
- Simplify: x = 100

These are the solutions to the given inequalities and equations.