anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
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?
amistre64
  • amistre64
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!!)
anonymous
  • anonymous
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?

Looking for something else?

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

More answers

amistre64
  • amistre64
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.
amistre64
  • amistre64
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 :)
amistre64
  • amistre64
3^(2/3) = (1/9)^((2/3) -1) cbrt(9) = (1/9)^(-1/3) cbrt(9) = 9 ^ (1/3) cbrt(9) = cbrt(9) (yessssss!!!!!)
amistre64
  • amistre64
(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)
amistre64
  • amistre64
forget the straggler there, hes being defiant to the end....
anonymous
  • anonymous
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:)
anonymous
  • anonymous
Qick question. Your writting them on both sides here (number four) simply to eliminate them on one of the side, correct?
amistre64
  • amistre64
yes

Looking for something else?

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