Post a New Question

Algebra 2

posted by .

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 -

    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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question