I can't figure out this math question (It is college algebra) Please help!?

Question

Q=rx^2c Solve for C So basically I need C=....?

Answer 1

Is the C part of the exponent? If so, you'll need to use logarithms...
If not, the other solution that several people gave you, is correct.
Also, you're using C and c to indicate apparently the same thing. This also adds to the confusion.
At any rate, if you mean Q = rx^(2c)
then (Q/r) = x^(2c)
and log base x (Q/r) = 2c so
(1/2) log base x (Q/r) = c
If you'd rather not use log base x you can use any other base such as ln or log base 10. Using ln,
ln (Q/r) = 2c ln x
so ln (Q/r)/(2 ln x) = c
which you'll recognise the change of base formula lurking in there.

Answer 2

I used the definition of a logarithm.

LOGaB = X (log with base a to B = x) is equivalent to

a^x = b (a raised to the x power = b)

Using this to solve for C:
Q = rx^2c
Q/r = x^2c (devide both sides by r)
then using the def of logarithms:
LOGx(Q/r) = 2c and deviding both sides by 2
c = (1/2)LOGx(Q/r)
This definition always tricks me. But in this, matching the symbols of the definition I gave, x is the equivalent of a, (Q/r) is the equivalent of B and 2c is the equivalent of x.

Answer 3

Q=rx^2C
dividing by rx^2
C=Q/rx^2

Answer 4

How about c=Q/rx^2?

Answer 5

Q = r(x^2)c

Divide both sides by rx^2

Q / (rx^2) = C

Answer 6

If you mean Q= rx^(2c), then c= ln(Q/r)/2lnx. However, if Q= r(x^2)c, then c= Q/r(x^2). I'm sure of these.

Answer 7

c= Q/(rx^2)

Answer 8

assuming that what you wrote is really Q=rx^(2c).....

I'll give you a hint, but I'm not solving it for you... use logarithms.

Answer 9

c = Q/(rx^2)

Answer 10

eighty 5 = 10* (log B- log 10^-sixteen) eighty 5 = 10*(log B + sixteen) eighty 5 = 10log B +one hundred sixty -75 = 10 log B log B = -7.5 B = 10^-7.5 i.e B = 10^-(15/2) B = 3.sixteen^ -15 (approx.) thats the respond i'm getting- seems such as you have been incorrect.