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magepker728 Group Title

1help step by step please on how to do this 5^x=4^x+1

  • one year ago
  • one year ago

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  1. JamesJ Group Title
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    Well, clearly x = 1 is a solution, yes?

    • one year ago
  2. ZeHanz Group Title
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    Did you mean: 5^x=4^(x+1)

    • one year ago
  3. magepker728 Group Title
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    yes

    • one year ago
  4. ZeHanz Group Title
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    The you could also write :\[5^x=4^{x+1} \Leftrightarrow 5^x=4^x \cdot 4^1 \]

    • one year ago
  5. JamesJ Group Title
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    yes to \[ 5^x = 4^{x+1} \] ?

    • one year ago
  6. magepker728 Group Title
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    yes @JamesJ

    • one year ago
  7. ZeHanz Group Title
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    Now you can divide by 4^x

    • one year ago
  8. ZeHanz Group Title
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    And use \[\frac{ a^x }{ b^x }=\left( \frac{ a }{ b } \right)^x\]

    • one year ago
  9. magepker728 Group Title
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    |dw:1359924788627:dw|

    • one year ago
  10. JamesJ Group Title
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    Alternatively, take the natural log of both sides and you have \[ \ln 5^x = \ln 4^{x+1} \] \[ x \ln 5 = (x+1) \ln 4 \] Now solve for x.

    • one year ago
  11. magepker728 Group Title
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    i got to this part

    • one year ago
  12. magepker728 Group Title
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    @JamesJ that is where my problem is solvinf for that x :D

    • one year ago
  13. JamesJ Group Title
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    So subtracting ln 4 . x from both sides, we have \[ (\ln 5 - \ln 4)x = \ln 4 \] Yes? Now take it from there ...

    • one year ago
  14. ZeHanz Group Title
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    I did the following: \[5^x=4^x \cdot 4 \Leftrightarrow \left( \frac{ 5 }{ 4 } \right)^x=4\]\[x=\log_{\frac{ 5 }{ 4 }} 4=\frac{ \ln4 }{ \ln \frac{ 5 }{ 4 } }=\frac{ \ln 4 }{ \ln5-\ln4 }\]

    • one year ago
  15. ZeHanz Group Title
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    @JamesJ: sorry, didn't mean to give it away...

    • one year ago
  16. magepker728 Group Title
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    thank you every1 for ur help

    • one year ago
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