I'm having really big problems with these specific Exponential Equations.
Please help. Calculations of HOW you do it would just simply be amazing. Ty.
Stacey Warren - Expert brainly.com
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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?
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.
3 * 4^(3-1) = 48
3 * 4^2 = 48
3 * 16 = 48
48 = 48 (yay!!)
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?
forget the straggler there, hes being defiant to the end....
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:)
Qick question. Your writting them on both sides here (number four) simply to eliminate them on one of the side, correct?