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dmartinez130 Group Title

Solve log4 32=x+2. Can someone please help me solve this?

  • 2 years ago
  • 2 years ago

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  1. dmartinez130 Group Title
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    4 is the base of log.

    • 2 years ago
  2. dmartinez130 Group Title
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    \[\log _{4}32=x+2\]

    • 2 years ago
  3. myininaya Group Title
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    To solve for x Subtract 2 on both sides

    • 2 years ago
  4. dmartinez130 Group Title
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    So, log4 30=x?

    • 2 years ago
  5. Ravus Group Title
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    Change it to the exponential form. Tell me what you see

    • 2 years ago
  6. Ravus Group Title
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    I think my way is easier :p try this first ^

    • 2 years ago
  7. Ravus Group Title
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    4^(x+2)=32 do you know how to get that?

    • 2 years ago
  8. dmartinez130 Group Title
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    No I don't :/

    • 2 years ago
  9. myininaya Group Title
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    \[\log_432=x+2\] -2 -2 \[\log_432-2=x\]

    • 2 years ago
  10. Ravus Group Title
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    The whole point of logs, is that they were "made" to "Solve for" an exponent. So if I had an expression that was 2^x=100, the only way to "find x" is to use logarithms. How? You can re-write any exponential equation to look like a logarithm, and vise versa.

    • 2 years ago
  11. Ravus Group Title
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    To go from log to exponential, you need to write the "answer" as the EXPONENT of the BASE. 4 is the base of the log, so write x+2 as the exponent of 4. 4^(x+2)

    • 2 years ago
  12. Ravus Group Title
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    and what does that equal? Well you only have one number left; the number you were taking the log of. 32. This is a basic skill you need to learn :) Writing logs as exponentials, and writiing exponentials as logs.

    • 2 years ago
  13. myininaya Group Title
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    Now if you want to simplify \[\log_432\] I would say let g equal that so we have \[g=\log_432\] writing in exponential form we get \[4^g=32\] g=3 doesn't work b/c 4(4)(4)=16(4)=64 g=3/2 doesn't work b/c (4)^(3/2)=2^3=8 g=5/2 works b/c (4)^(5/2)=2^5=32 Booyah! :) So that means \[\log_4(32)=\frac{5}{2}\] So what does that mean x equals or simplifies to?

    • 2 years ago
  14. myininaya Group Title
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    You could just use your calculator though instead of that method right there sometimes that method will be too difficult to use

    • 2 years ago
  15. myininaya Group Title
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    A good formula to remember is the change of base formula \[\frac{\ln(a)}{\ln(b)}=\log_b(a)\]

    • 2 years ago
  16. Ravus Group Title
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    \[4^{x+2}=32 \] So how do you solve this? There's a trick you'll probably have to often use: you need express both numbers as the same base! You can probably tell that 4 is simply 2^2, and 32 is 2^5 That's what I mean by "the same base" So once you do that, you've got this: \[2^{2(x+2)}=2^{5}\] Once you have something looking like that, you can solve for x! You simply "ignore" the "big" 2s, and you get 2(x+2)=5 solve for x now, and you get 0.5, or 1/2 :)

    • 2 years ago
  17. myininaya Group Title
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    using my way we get \[\frac{5}{2}-2=x\]

    • 2 years ago
  18. myininaya Group Title
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    Remember since \[\log_4(32)=\frac{5}{2}\]

    • 2 years ago
  19. Ravus Group Title
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    The trick to these is: 1. writing it in exponential form (learn to do this!! you'll use it all the time) 2. Expressing both sides of your new (but equilivant to the log expression) equation with the same bases. 3."zooming in" on the exponents, ignoring the "big numbers" (which are now the same) and solving for x like a regular algebra problem.

    • 2 years ago
  20. Ravus Group Title
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    myininaya i like your solution, it's just that I'm close to the asker's age, I think, and I remember that this is the particular way they want these questions done, even if it is a drag , lol

    • 2 years ago
  21. Ravus Group Title
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    they=the teachers

    • 2 years ago
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