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Homework Help: Solving logarithmic and exponential equations

  1. Oct 15, 2013 #1
    1. The problem statement, all variables and given/known data
    I'm trying to solve trying logarithmic and exponential equations but for the life of me I can't seem to figure out what I'm doing wrong.

    2. Relevant equations
    Log equation: log3(x+2)+1 = log3x²+4x
    Exp equation: 2x + 2x+1 = 3

    3. The attempt at a solution
    For the first one, here's what I did:
    • 1 = log3(x²+4x) - log3(x+2)
    • log3((x²+4x)/(x+2)) = 1
    • log3((x² +4x)/(x+2)) = 1
    • log3((x+4x)/2) = 1 --Can I do this? Dividing the denominator and numerator by the same value within a log
    • log3(x+2x) = 1
    • log3(3x) = 1
    • log33 + log3x = 1
    • 1 + log3x = 1
    • log3x = 0
    • x = 1

    And then for the second one:
    • 2x + 2x+1 = 3
    • I figured I'd put them all on the same base so I tried to find 2 to what power equals 3
    • So I did ln(3)/ln(2) = 1.584962501, which for simplicity's sake I'll replace by y here
    • And when I put in 2y in my calculator it does give me 3 so it must be correct
    • x + x + 1 = y
    • 2x + 1 = y
    • x = (y-1)/2
    But when I put the result back in the original equations, none of the answers I found seem to be correct.
    I'm at my wit's end here, I don't know what else to try.

    Any help is greatly appreciated.
  2. jcsd
  3. Oct 15, 2013 #2


    Staff: Mentor

    For the log expression on the right side, do you mean log3(x2 + 4x)? That seems to be what you're doing in your work below.
    • I have no idea what you did. How did you go from (x2 + 4x)/(x + 2) to (x + 4x)/2? That's not a valid step.
    • Where did this (above) come from?
    Last edited: Oct 15, 2013
  4. Oct 15, 2013 #3


    User Avatar
    Homework Helper

    That should be written: log3(x+2)+1 = log3(x²+4x)

    Parentheses are important!

    • OK so far.

      You can divide a rational expression within a log argument. But your algebra is totally wrong here, because ##\frac{x^2+4x}{x+2} \neq \frac{x+4x}{2}##. Please review your algebra.

      Instead of this, I would just take antilogs of both sides to get:

      ##\frac{x^2+4x}{x+2} = 3##

      and solve that quadratic. Be careful about which solution(s) is/are admissible.

      • Doesn't help because ##a^x + a^y = a^z## does NOT imply that ##x + y = z##. In fact, there's no "easy" relationship between x, y and z here.

        To solve, observe that ##2^{x+1} = 2.2^x## (the dot signifies multiplication). Now put ##2^x = y## to get a simple linear equation.
  5. Oct 15, 2013 #4
    Thanks, I didn't think of using an antilog, solved it smoothly.

    I kind of don't follow what happened here, where did 2x go?
  6. Oct 15, 2013 #5


    Staff: Mentor

    It's still there.
    2x + 1 = 2 * 2x
  7. Oct 15, 2013 #6
    Yeah but I mean the original formula was 2x + 2x+1 = 3, you moved it to the other side of the equals sign and it became 2 times 2x. Why not 3-2x?

    I'm sure it's obvious and I completely missed it but it's 1am here and trying to finish this before going to sleep
  8. Oct 15, 2013 #7
    The quantity log3((x² +4x)/(x+2)) represents the power to which you have to raise the number 3 to get ((x² +4x)/(x+2)). The equation log3((x² +4x)/(x+2))=1 says that the number 1 is the power to which you have to raise the number 3 to get ((x² +4x)/(x+2)). So, 31=((x² +4x)/(x+2))

  9. Oct 15, 2013 #8


    Staff: Mentor

    Because doing that won't help you. If you subtract 2x from both sides, you get
    2x + 1 = 3 - 2x
    This isn't wrong, but it's not helpful, either, in getting you closer to a solution.

    This is what you need to do:
    2x + 2x+1 = 3
    ==>2x + 2*2x = 3
    ==> 2x(1 + 2) = 3
    ==>2x * 3 = 3
    ==>2x = 1
    Now you can solve for x.
  10. Oct 15, 2013 #9
    Aaaah right 2x+1 = 2*2x is the same as 21*2x which makes the rest distributable now

    Great thanks I can finally go catch some Zs now, you've been a real help!
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