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gioanny39x

  • 3 years ago

What is the solution of log 2x – 3^125 = 3 ? x = 1 over 3 x = 1 x = 7 over 3 x = 4

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  1. telltoamit
    • 3 years ago
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    log basex a = b implies x^a = b

  2. gioanny39x
    • 3 years ago
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    i dont understand

  3. Wired
    • 3 years ago
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    Is this correct? \[\Large \log (2x-3^{125}) = 3\]

  4. telltoamit
    • 3 years ago
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    nah the base is 2x - 3

  5. gioanny39x
    • 3 years ago
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    so confuse >.<

  6. Wired
    • 3 years ago
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    @telltoamit Only @gioanny39x can say that for sure. Gio, you need to confirm if I have the equation correct or not. Also, your formula is wrong. \[\Large log_B(X) = Y\] is the same as \[\Large X = B^{Y}\] Example: \[\Large log_10(1000) = 3\] B = 10 X = 1000 Y = 3 Using the 2nd form of the equation: \[\Large X = B^{Y}\] \[\Large 1000 = 10^{3}\]

  7. ash2326
    • 3 years ago
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    \[\log_{2x-3} 125=3\] Is this your question @gioanny39x ?

  8. telltoamit
    • 3 years ago
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    only this gives the answer in options lol

  9. gioanny39x
    • 3 years ago
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    |dw:1343929733059:dw|

  10. Wired
    • 3 years ago
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    Ok, so: B = 2D-3 (easier to use D to not confuse with X) X = 125 Y = 3 \[\Large X = B^{Y}\] \[\Large 125 = (2D-3)^{3}\] Solve for D. Hint: 125 is the cube of a single digit number.

  11. gioanny39x
    • 3 years ago
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    (2d-3)(2d-3)(2d-3) 4d^2-6d-6d+9 4d^2-12d+9(2d-3) 8d^3-12d^2 -24d^2+36d 28d-27 125=8d^3=36d^2+64d+27?

  12. gioanny39x
    • 3 years ago
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    @Wired

  13. Wired
    • 3 years ago
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    You're creating too much work for yourself. What is the cube root of 5? Hint: \[\Large X^{3} = (15Y+5)^{3}\] \[\Large X = (15Y+5)\] If you have two items that when cubed equal each other, the items equal each other.

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