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anonymous

  • one year ago

Solve 3(2x) = 7(x−1).

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  1. nincompoop
    • one year ago
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    what do you think you should do first?

  2. anonymous
    • one year ago
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    its x=7

  3. nincompoop
    • one year ago
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    you can deal with distributive property first or you can settle the coefficients by making the right hand side and left hand side have the same value of coefficients

  4. nincompoop
    • one year ago
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    please do not give out answers

  5. anonymous
    • one year ago
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    but if I do the substitution method wouldn't I need to do something else first? @nincompoop

  6. nincompoop
    • one year ago
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    substitution method how?

  7. anonymous
    • one year ago
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    its more like 3^(2x) = 7^ (x-1)

  8. nincompoop
    • one year ago
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    ohhh

  9. nincompoop
    • one year ago
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    hahaha

  10. nincompoop
    • one year ago
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    did you learn about logarithms yet?

  11. anonymous
    • one year ago
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    yes but it makes no sense

  12. nincompoop
    • one year ago
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    laughing out loud

  13. anonymous
    • one year ago
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    this is serious lol

  14. anonymous
    • one year ago
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    I need this class

  15. anonymous
    • one year ago
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    Yeah nincompoop.

  16. nincompoop
    • one year ago
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    use this \(\large log(a^b) = b~log(a) \)

  17. nincompoop
    • one year ago
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    |dw:1437603054834:dw|

  18. anonymous
    • one year ago
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    so I multiply them?

  19. nincompoop
    • one year ago
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    no do the log first

  20. anonymous
    • one year ago
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    so the logs are in front of the numbers?

  21. nincompoop
    • one year ago
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    ye

  22. nincompoop
    • one year ago
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    please read this page then proceed with your problem later http://www.mathsisfun.com/algebra/exponents-logarithms.html

  23. anonymous
    • one year ago
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    will you still help me

  24. nincompoop
    • one year ago
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    maybe, but read and take notes first

  25. anonymous
    • one year ago
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    okay

  26. freckles
    • one year ago
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    Here is an example: \[5^x=6^{2x+3} \\ \text{ take } \log( ) \text{ of both sides } \\ \log(5^x)=\log(6^{2x+3}) \\ \text{ we do this so we can use } \log(r^x)=x \log(r) \text{ this is called power rule } \\ \text{ so bringing the powers down } \\ x \log(5)=(2x+3) \log(6) \\ \text{ distribute on the right hand side } \\ x \log(5)=2x \log(6)+3 \log(6) \\ \text{ now collect your terms with } x \text{ on one side } \\ x \log(5)-2x \log(6)=3 \log(6) \text{ I just subtracted } 2x \log(6) \text{ on both sides } \\ \text{ now we are going to factor the } x \text{ out on the left } \\ x(\log(5)-2 \log(6))=3 \log(6) \\ \text{ now we are going to divide both sides by what } x \\ \text{ is being multiplied by } \\ x=\frac{3 \log(6)}{\log(5)-2 \log(6) }\] Can you show what you have doing just this step: so you try taking the log of both sides and bringing down the powers. ?

  27. anonymous
    • one year ago
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    omgoodness thank you so much

  28. freckles
    • one year ago
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    Did you want to show me the one step I was mentioning? Like taking log of both sides then bringing down the powers as a result of using the power rule for log?

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