A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

EXPONENTIAL EQUATIONS... I'm having really big problems with these specific Exponential Equations. 2^x=1/32 3⋅4^(x-1)=48 3^x=(1/9)^(x-1) 4^x×2^x=64 Please help. Calculations of HOW you do it would just simply be amazing. Ty.

  • This Question is Closed
  1. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    2^x = 1/32 what it is asking is what value of "x" will make this a true statement. You need to know some basic information rules here: B^x = y is the stated problem logB(B^x) = logB(y) take the logB of both sides; x logB(B) = logB(y) the "x" is an exponent and logs say that you can pull it out and multiply it; which is what I did. x = logB(y) ...logB(B) = 1 x(1) = x. lets solve your first equation: 2^x = 1/32 log2(2^x) = log2(1/32) x (log2(2)) = log2(1/32) x = log2(1/32) this is good; but we can manipulate it some more based on rules of logarithms. x= log2(1) - log2(32) x = 0 - 5 x = -5 if I kept track of everything :) a negative exponent is like a reciprocal. 2^-5 equals 1/2^5 which equals 1/32 does that make sense?

  2. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Lets see how that works for your next one: 3⋅4^(x-1)=48 ...divide by 3 4^(x-1) = 48/3 = 16 If you know that 4^2 = 16 then it is simple: x-1 has to equal 2. x=3 3 * 4^(3-1) = 48 3 * 4^2 = 48 3 * 16 = 48 48 = 48 (yay!!)

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    This makes a lot of sense, and the second one i understand compelty. But the first one, well, i'm not entirely sure why you use "log" in the first one?

  4. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    exponents and logs are inverses of each other; and they really have the same rules. logs make working with exponents easier because it gives us tools to manipulate them with.

  5. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    3^x=(1/9)^(x-1) x = log3(1/9^(x-1)) x = (x-1) (log3(1/9)) 3^? = 1/9; 3^? = 1/(3^2) 3^? = 3^-2 log3(1/9) = -2 x = (x-1)(-2) x = -2x + 2 x + 2x = 2 3x = 2 x = 2/3 might wanna double check that :)

  6. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    3^(2/3) = (1/9)^((2/3) -1) cbrt(9) = (1/9)^(-1/3) cbrt(9) = 9 ^ (1/3) cbrt(9) = cbrt(9) (yessssss!!!!!)

  7. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    (4^x) (2^x) = 64 log2((4^x)(2^x)) = log2(64) = 6 log2(4^x) + log2(2^x) = 6 x(log2(4)) + x(log2(2)) = 6 x(2+1) = 6 x= 6/(2+1) = 2 (4^2) (2^2) = 64 (16)(4) = 64 64 = 64 ...... got it :) x (log)

  8. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    forget the straggler there, hes being defiant to the end....

  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Wow, Thanks mate. Made things much more clear. I'm not entirely confident about 3 and 4, but I'm starting to understand the concept of these problems, i'll just read through the explanations a few times, i'm sure i'll eventually understand it. Thanks again:)

  10. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Qick question. Your writting them on both sides here (number four) simply to eliminate them on one of the side, correct?

  11. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  12. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.