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Exponential and logarithmic Equation Problems

  1. May 11, 2017 #1
    1. The problem statement, all variables and given/known data

    Evaluate each of the following expressions without using a calculator.

    1) log216√8


    Solve for the unknown value in each of the following equations without using a calculator.

    2) 3(x+4)−5(3x)=684

    3) 7(42x)=28(4x)

    2. Relevant equations

    Exponent law for multiplication

    3. The attempt at a solution

    1)

    log2(16)√8
    =log232√2
    =log2321/2

    Im not really sure what to do at this point or if my approach is just completely wrong

    2)

    3(x+4)−5(3x)=684
    =(3x)(34)-5(3x)=684
    Let a = 3x
    (a)(34)-5a=684

    Once again Im not sure if my approach is completely wrong or Im just at a loss on what to do afterward.

    3)

    7(42x)=28(4x)
    7(4x)2=28(4x)
    Let a=4x
    7a2=28a
    a2=28a/7
    a2=4a

    Im not sure what to do after this
     
  2. jcsd
  3. May 11, 2017 #2
    ##\log_2 ((16)\sqrt{8})## or ##\sqrt{8}\times \log_2 ((16))## ?
     
  4. May 11, 2017 #3
    The former

    I edited the OP to make it a little clearer
     
  5. May 11, 2017 #4
    Use ##\log_a(bc) = \log_a(b) + \log_a(c)##, ##16 = 2^4## and same for ##\sqrt{8}##.

    For the other two you need to take log on both sides.
     
  6. May 11, 2017 #5
    so the first one would be:

    ##\log_2(16) + \log_2(\sqrt{8})##
    ##\log_2(2^4) + \log_2(2\sqrt{2})##

    and Im not really sure what to do with the square root

    For the other two questions, is what I have so far correct and I need to take the log on both sides from the point where Im at or is the whole thing wrong?
     
  7. May 11, 2017 #6
    No you are not getting it. You should try to use ##\log_a(b^c) =c\log_ab##. Now if ##a =b## then ##c\log_a(a) = ?##

    What you did is correct, now you need to solve for 'a' then take log to 'x'.
     
  8. May 11, 2017 #7

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Write ##\sqrt{8}## in form of 2a, power of 2 .
     
    Last edited by a moderator: May 12, 2017
  9. May 12, 2017 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

    How much is 34?

    Rewrite the equation in form a2-4a =0, and factor out a. Remember, a product is zero if any of the factors is zero. What solutions do you get for a? Are both valid?
     
  10. May 19, 2017 #9
    Sorry for the super late reply. I decided to continue on with other questions and the next unit before coming back to this.
    1)

    ##\log_2(16) + \log_2(\sqrt{8})##
    ##\log_2(2^4)(2^3)^{0.5}##
    ##\log_2(2^4)(2^{1.5})##
    ##\log_2(2^{5.5})##
    ##5.5\log_2(2)##
    ##=5.5##

    2)

    wow I dont know why this question shook me. The answer was in my face the entire time haha.
    (starting from where I left off in the OP)
    ##81a-5a=684##
    ##76a=684##
    ##a=684/76##
    ##a=9##
    ##3^x=9##
    ##x=2##

    3)
    (starting from where I left off in the OP)
    ##a^2=4a##
    ##4^{2x}=(4)4^x##
    ##4^{2x}=4^{x+1}##
    ##2x=x+1##
    ##2x-x=1##
    ##x=1##

    EDIT: thanks for the heads up on the typo
     
    Last edited: May 19, 2017
  11. May 19, 2017 #10
    I guess it is correct except one typo.
     
  12. May 19, 2017 #11

    Mark44

    Staff: Mentor

    It's a good habit to get into to check your answer by substituting your value of x in the original equation.
    You started off in the direction that @ehild suggested, but then went off in a different direction. Following the direction she suggested you would have this:
    ##4^{2x} - 4\cdot4^x = 0##
    ##4^x(4^x - 4) = 0##
    Are you sure there aren't solutions other than x = 1, which is the one you showed?
     
  13. May 19, 2017 #12
    doing it that way I get:

    ##a^2-4a=0##
    ##a(a-4)=0##
    ##a=4## or ##a=0##

    ##a=4##
    ##4^x=4##
    ##x=1##

    ##a=0##
    ##4^x=0##
    *no solutions*

    therefore the correct answer is ##x=1##
     
  14. May 19, 2017 #13

    Mark44

    Staff: Mentor

    Looks good!
     
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