Algebra 2
posted by Linda on .
How do I solve these inequalities:
e^3x>e^x1
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^x1
you probably meant:
e^(3x) > e^(x1), (huge difference from what you typed)
ln both sides
ln (e^(3x)) > ln e^(x1)
3x > x1
2x > 1
x > 1/2
PROOF:
http://www.wolframalpha.com/input/?i=plot+y+%3D+e%5E%283x%29+%2C+y+%3D+e%5E%28x1%29+from+1+to+0
6log (subscript of 4) x > 24
divide both sides by 6
log_{4} 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 = log_{4} 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