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rosedewittbukater

  • 3 years ago

Algebra 2 help and explanation? I get confused when the base of the log isn't 10. Solve the equation and round to the nearest ten-thousandth. log2 4x = 5 (2 is the base)

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  1. rosedewittbukater
    • 3 years ago
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    And I need help with 2 log3 x = 54

  2. whpalmer4
    • 3 years ago
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    \[log_2{4x} = 5\] Raise both sides to the 5th power of the log base to get rid of the log: \[4x = 2^5\] \[x= 32/4 = 8\] Check our work: \[log_2{(4*8)} = 5\] \[2^5 = 32 = 4*8\] Remember the logarithm base n of x is just the number to which you raise n to get x. So, the log base 2 of 32 is 5, because 2^5 = 32.

  3. whpalmer4
    • 3 years ago
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    \[2\log_3 x = 54\] I'd divide both sides by 2 to get \[\log_3 x = 27\] Can you tell me what x is now?

  4. whpalmer4
    • 3 years ago
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    Or is the equation \[2\log{(3x)}=54\]?

  5. whpalmer4
    • 3 years ago
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    I'm suspicious that the second problem isn't correct...

  6. rosedewittbukater
    • 3 years ago
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    @whpalmer4 thanks so much! The second one has a base of 3.

  7. rosedewittbukater
    • 3 years ago
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    for the second one I changed it to log3 x^2 = 54 and then 3^54 = x^2 3^27 = x x = 7.625 times 10^12

  8. whpalmer4
    • 3 years ago
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    I would check your rounding carefully on that last one, you've gotten the right number, it would be a shame to still get it wrong (I'm not taking a position either way)

  9. musiclover101
    • 3 years ago
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    it would be|dw:1357961835423:dw|

  10. musiclover101
    • 3 years ago
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    and so the answer is 8 if you solve for x

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