Question

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.

Answer 1

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?

Answer 2

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!!)

Answer 3

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?

Answer 4

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.

Answer 5

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 :)

Answer 6

3^(2/3) = (1/9)^((2/3) -1)

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

Answer 7

(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)

Answer 8

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

Answer 9

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:)

Answer 10

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

Answer 11

yes