Posted by **Linda** on Thursday, April 17, 2014 at 8:45pm.

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?

- Algebra 2 -
**Reiny**, Thursday, April 17, 2014 at 9:15pm
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

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

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