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Tricky trignometry

  1. Jul 15, 2006 #1
    i got 2 problems that are really hard to solve. i dont know wether they are trick question or they just can be incorperated in a formula.i have taken days to solve them but in vain. please help!

    a) 10(3^(2x+1)) = 2^(4x-3)

    b) (2/3x)^(log2) = (9x)^(log3)

    thanks you very much:tongue:
  2. jcsd
  3. Jul 15, 2006 #2
    Wait, this is just solve for x, right? Trigonometry?

    I can help, but what have you done so far? Have you used all the properties of logs?
  4. Jul 15, 2006 #3
    What trignometry in these problems
  5. Jul 15, 2006 #4


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    Hello vijay,

    this problem is fairly hard to solve with trigonometry :wink:

    Have you tried the following identities?





  6. Jul 15, 2006 #5
    yes, i have tried all of them, they dont seem to work!
  7. Jul 15, 2006 #6
    yes, the rpoblem is trignometric, but all my efforts fail.
  8. Jul 15, 2006 #7
    and for info. this sum is a non calculator one....so....
    anyway..for the first question, it might be helpful to expand out 10 into 5*2(accoring to me). the brought the 2 to the other side and the after further simplying, found out that (16/9)^x = 240....i dont know how to go on without using a calc. hope some1 has a better solution to the problem.
  9. Jul 15, 2006 #8


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    Let's take a look at problem a)



    Can you apply the second and third identity I posted earlier here?


  10. Jul 15, 2006 #9
    how would that help. the equation you have given is just the manipulated version of the problem. i tried to simplfy it, but it doesnt do any good.i come back to the same question.
  11. Jul 15, 2006 #10


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    Ok, so I have to ask what you want to achieve actually. :smile: Do you need an expression for x or something else?


  12. Jul 15, 2006 #11
    i would like to simplfy the rpbolem, making x the subject, were the whole equation is representated in exponential format, were x is equal to something in exponential form were there are no logarithms or any variable present.
  13. Jul 15, 2006 #12


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    I'll try to show the next steps I'd take:



    Can you solve it for x?


  14. Jul 15, 2006 #13
    yes yes..thats exactly wut i did...but now were near solution...lol...
  15. Jul 15, 2006 #14
    the log 10 is 1 actually.....
  16. Jul 15, 2006 #15


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    Hang on...what makes you so sure that the base was chosen to be 10? :wink:

    What do you mean with nowhere near the solution? Aren't these steps taking you closer and closer to it :confused:

    You've tried this step yet?




    Last edited: Jul 15, 2006
  17. Jul 15, 2006 #16


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    Ok, the simple way is to simplify the equation you get, then separate the unknown x into one side, and solve the equation from there.
    I'll start off the first problem for you.
    [tex]10 \times 3 ^ {2x + 1} = 2 ^ {4x - 3}[/tex]
    This looks too complicated, we must then use the following formulae to simplify it a bit:
    ab + c = ab ac
    ab - c = ab / ac
    (ab)c = abc
    (a / b)c = ac / bc
    We have:
    [tex]\Leftrightarrow 10 \times 3 ^ {2x} \times 3 = \frac{2 ^ {4x}}{2 ^ 3}[/tex]
    [tex]\Leftrightarrow 30 \times {(3 ^ {2})} ^ x = \frac{{(2 ^ {4})} ^ x}{8}[/tex]
    [tex]\Leftrightarrow 30 \times 9 ^ x = \frac{16 ^ x}{8}[/tex]
    This looks much less complicated, right?
    So, let's isolate x:
    [tex]\Leftrightarrow \frac{9 ^ x}{16 ^ x} = \frac{1}{8 \times 30}[/tex]

    [tex]\Leftrightarrow \left( \frac{9}{16} \right) ^ x = \frac{1}{240}[/tex]
    Now if ax = b, then how can you solve x in terms of a, and b?
    Hint: Use logarithm. :)
    The second one can be done exactly in the same way.
    Can you go from here? :)
    By the way, it's certainly not a trigonometry problem. :)
    Last edited: Jul 15, 2006
  18. Jul 15, 2006 #17


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    Very nice!

    Vijay, if you'd like to compare VietDao's solution with mine, here is the result I get:


    Both ways should give the same values for x.


  19. Jul 15, 2006 #18
    lol...dats wut i got in the first place!!!!!!!!!
  20. Jul 15, 2006 #19
    how do you simply dat?.....dats the problem....i got the exact same ans.
  21. Jul 15, 2006 #20
    lol....sry...not trignometry....logarithms....
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