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u0860867

  • 2 years ago

help needed

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  1. u0860867
    • 2 years ago
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  2. UnkleRhaukus
    • 2 years ago
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    \[\log_7M=\frac{\log_910-\log_95}{\log_97}\]

  3. UnkleRhaukus
    • 2 years ago
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    first use this log law in the numerator of the right hand side \[\boxed{\log_b x-\log_by=\log_b\frac xy}\]

  4. u0860867
    • 2 years ago
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    yeah i did that so i got log 2

  5. UnkleRhaukus
    • 2 years ago
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    now use the change of base formula \[\large\boxed{\log_b x=\frac{\log_c x}{ \log_c b}}\]

  6. u0860867
    • 2 years ago
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    0.31 for the numerator

  7. UnkleRhaukus
    • 2 years ago
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    leave it in log form

  8. u0860867
    • 2 years ago
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    ok

  9. UnkleRhaukus
    • 2 years ago
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    So you have \[\log_7M=\frac{\log_92}{\log_97}\] comparing with the change of base formula \[\boxed{\log_b x=\dfrac{\log_c x}{ \log_c b}}\] we see that \(M= ~ . . .\)

  10. u0860867
    • 2 years ago
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    so will u get Log7M= 0.35 @UnkleRhaukus

  11. u0860867
    • 2 years ago
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    so will M =2 ?????

  12. UnkleRhaukus
    • 2 years ago
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    dont convert to decimal

  13. UnkleRhaukus
    • 2 years ago
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    yeah that's it M=2

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