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Thread: Indices Equations Questions

  1. #1
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    Indices Equations Questions

    Hello there Math master, (I wonder if I could become one later...)

    I have this several questions in which I could not find the answer..I'm stuck.

    If $\displaystyle 10^m = 2$ and $\displaystyle 10^n = 3$, find the value of 10^3m-n

    Now I have tried randomly and find out that m = 1/5, but I do not know the proper way how to find m and n respectively. The book said that we need to find the common factor of 10 and 2, but so far I could not find any because the number seems very weird...

    There's also this another question which is nearly the same:

    Given that $\displaystyle 3^x = 5$ and $\displaystyle 3^y = 7$, find the value of 3^2x-3y, giving your answer in fraction.

    Thank you for your kind help.
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  2. #2
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    $\displaystyle 10^m = 2$ so cubing $\displaystyle (10^m)^3 = 10^{3m} = 2^3 = 8$.

    So $\displaystyle 10^{3m-n}={10^{3m} \over 10^n}={8\over3}$.
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  3. #3
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    Quote Originally Posted by maddas View Post
    $\displaystyle 10^m = 2$ so cubing $\displaystyle (10^m)^3 = 10^{3m} = 2^3 = 8$.

    So $\displaystyle 10^{3m-n}={10^{3m} \over 10^n}={8\over3}$.
    Thank you for your time answering my question but there are several things I do not comprehend. You're cubing the 10^m but why you're not cubing the 10^n too? Could you explain it a little bit step-by-step?
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  4. #4
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    Because you were asked for $\displaystyle 10^{3m-n}$, and not $\displaystyle 10^{3m-3n}$. Perhaps it would be clearer if I did...

    $\displaystyle 10^{3m-n} = {10^{3m}\over 10^n} ={10^{m+m+m}\over10^n}= {10^m\cdot10^m\cdot10^m\over10^n} = {2\cdot2\cdot2\over3}=\frac83$

    ?
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  5. #5
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    Quote Originally Posted by maddas View Post
    Because you were asked for $\displaystyle 10^{3m-n}$, and not $\displaystyle 10^{3m-3n}$. Perhaps it would be clearer if I did...

    $\displaystyle 10^{3m-n} = {10^{3m}\over 10^n} ={10^{m+m+m}\over10^n}= {10^m\cdot10^m\cdot10^m\over10^n} = {2\cdot2\cdot2\over3}=\frac83$

    ?
    Ok I got it, so would this be correct: (the second question I posted in the first post)

    square 3^x = 5 so that we got 3^2x = 25
    cube 3^y = 7 so that we got 3^3y = 343

    ultimately we have 25/344?
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  6. #6
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    Quote Originally Posted by Lites View Post
    Ok I got it, so would this be correct: (the second question I posted in the first post)

    square 3^x = 5 so that we got 3^2x = 25
    cube 3^y = 7 so that we got 3^3y = 343

    ultimately we have 25/344 may be a typo!
    yes. $\displaystyle \frac{25}{343}$
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