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anonymous
 5 years ago
EXPONENTIAL EQUATIONS...
I'm having really big problems with these specific Exponential Equations.
2^x=1/32
3⋅4^(x1)=48
3^x=(1/9)^(x1)
4^x×2^x=64
Please help. Calculations of HOW you do it would just simply be amazing. Ty.
anonymous
 5 years ago
EXPONENTIAL EQUATIONS... I'm having really big problems with these specific Exponential Equations. 2^x=1/32 3⋅4^(x1)=48 3^x=(1/9)^(x1) 4^x×2^x=64 Please help. Calculations of HOW you do it would just simply be amazing. Ty.

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02^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
 5 years ago
Best ResponseYou've already chosen the best response.0Lets see how that works for your next one: 3⋅4^(x1)=48 ...divide by 3 4^(x1) = 48/3 = 16 If you know that 4^2 = 16 then it is simple: x1 has to equal 2. x=3 3 * 4^(31) = 48 3 * 4^2 = 48 3 * 16 = 48 48 = 48 (yay!!)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This 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?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0exponents 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
 5 years ago
Best ResponseYou've already chosen the best response.03^x=(1/9)^(x1) x = log3(1/9^(x1)) x = (x1) (log3(1/9)) 3^? = 1/9; 3^? = 1/(3^2) 3^? = 3^2 log3(1/9) = 2 x = (x1)(2) x = 2x + 2 x + 2x = 2 3x = 2 x = 2/3 might wanna double check that :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.03^(2/3) = (1/9)^((2/3) 1) cbrt(9) = (1/9)^(1/3) cbrt(9) = 9 ^ (1/3) cbrt(9) = cbrt(9) (yessssss!!!!!)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.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)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0forget the straggler there, hes being defiant to the end....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wow, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Qick question. Your writting them on both sides here (number four) simply to eliminate them on one of the side, correct?
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