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Math Help - X variable in exponent, solve for x

  1. #1
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    X variable in exponent, solve for x

    Attached is the equation. Could someone show me the steps? I know you had to cross-multiply something.

    Is it x = 2x - 1?
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  2. #2
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    Quote Originally Posted by mwok View Post
    Attached is the equation. Could someone show me the steps? I know you had to cross-multiply something.

    Is it x = 2x - 1?
    NO! x is not equal to 2x - 1. Exponents are only equal to each other if the bases are the same.

    Remember that 16 = 2^4 and 8 = 2^3
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  3. #3
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    Please get the concept of "cross multiply" out of your life. It is not your friend. The fact that it doesn't even remotely apply to this problem suggests to me that it is as confusing as I like to say it is.

    You almost have this one, but failed to spot that the bases were different. If they were both 8 or both 16, you would be right on. You must change to a common base.

    16 = 2^{4} and 8 = 2^{3}

    Use those hints and show us what you get.
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    Ahh, thanks!
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  5. #5
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    How would I do this one since it can't be made into a common base?
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  6. #6
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    Quote Originally Posted by mwok View Post
    How would I do this one since it can't be made into a common base?
    In this case, you have to know that e^{\ln{(x)}} = x or \ln{(e^x)} = x because \ln{(x)} and e^x are inverses.
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    Quote Originally Posted by Chop Suey View Post
    In this case, you have to know that e^{\ln{(x)}} = x or \ln{(e^x)} = x because \ln{(x)} and e^x are inverses.

    Still don't understand. Can you solve the problem so I can see how it works?
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  8. #8
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    Quote Originally Posted by mwok View Post
    How would I do this one since it can't be made into a common base?
    Its actually a problem that is very similar to the first one. Whenever you see a power on one side and some number on the other, always take logs on both sides. And then try to play with the logs.

    For example:

    16^x = 8^{2x-1} \Rightarrow x \log 16 = (2x - 1) \log 8 \Rightarrow x (4\log 2) = (2x - 1) (3\log 2) \Rightarrow 4x = 3(2x-1)


    Similarly:

    e^{\ln (x+2)} = 6 \Rightarrow \ln e^{\ln (x+2)} = \ln 6 \Rightarrow  \ln (x+2) \ln e = \ln 6 \Rightarrow \ln (x+2) = \ln 6 \Rightarrow x+2 = 6

    This method is very mechanical. Identifying both sides as powers of the same base is faster, if you get used to it. But you will need practice before you can do that.
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    Ah, Thanks!
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  10. #10
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    Quote Originally Posted by mwok View Post
    Ah, Thanks!
    You are welcome... however do not ignore what Chop Suey said...

    You could have got the answer in one step! Did you observe that e^{\ln x} = x?

    Then e^{\ln (x+2)} = 6 \Rightarrow x+2 = 6
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  11. #11
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    Hmm, my solution is different from yours.

    What happened to the ln(e)?
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  12. #12
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    Quote Originally Posted by mwok View Post
    Hmm, my solution is different from yours.

    What happened to the ln(e)?
    Well ln stands for natural logarithm. Thus by definition, ln e = 1.
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  13. #13
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    Quote Originally Posted by Isomorphism View Post
    Well ln stands for natural algorithm. Thus by definition, ln e = 1.
    Okay but how did you get rid of the ln(x+2) and ln(6) so it becomes x+2=6? Did you just cancel ln from both sides?
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  14. #14
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    Quote Originally Posted by mwok View Post
    Okay but how did you get rid of the ln(x+2) and ln(6) so it becomes x+2=6? Did you just cancel ln from both sides?
    Oops I actually used what Chop Suey pointed out . I used the fact that ln has a unique inverse and thus \ln x = \ln y \Rightarrow x = y

    On second thought, use what Chop Suey suggested.... Use e^{\ln x} = x
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  15. #15
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    Quote Originally Posted by Isomorphism View Post
    Oops I actually used what Chop Suey pointed out . I used the fact that ln has a unique inverse and thus \ln x = \ln y \Rightarrow x = y

    On second thought, use what Chop Suey suggested.... Use e^{\ln x} = x

    Can you solve it using that method so I can see? I don't even know where the 2 and 6 goes.
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