
solving for unknowns
1.) If,
{4x^2  sqrt[3]{x^2} + 3}/{sqrt[6]{x^5}} = 4x^a  x^b + 3x^c}
then a= ______ , b= _______ , and c= _______ .
2.) If V=4r^2h and r/(Hh)= R/H, write V as a function of R
V= _______________
3) Solve for
S= (a^n) 1 /i
a= __________
4.) Solve for n:
P= R[(1x^n)/i]
n=___________

I'll give you a headstart on the first one ...
Multiply through by that for a start, and use fractional indices to make it easier to read:
Then equate the terms with coefficients 4, 1 and 3 and remember that when multiplying two terms you add their indices.

1.) If,
{4x^2  sqrt[3]{x^2} + 3}/{sqrt[6]{x^5}} = 4x^a  x^b + 3x^c}
then a= ______ , b= _______ , and c= _______ .
Does sqrt[3]{x^2] mean cuberoot of x^2? And sqrt[6](x^5) the 6th root of x^5?
If yes, then,
Rewriting your original equation,
[4x^2 x^(2/3) +3] / [x^(5/6] = 4x^a x^b +3x^c
4x^(2 5/6) x^(2/3 5/6) +3x^(5/6) = 4x^a x^b +3x^c
4x^(7/6) x^(1/6) +3x^(5/6) = 4x^a x^b +3x^c
Therefore, a = 7/6; b = 1/6; and c = 5/6 answer.

2.) If V=4r^2h and r/(Hh)= R/H, write V as a function of R
V= _______________
The idea here is to put V and R into one equation only.
We see that h is common in the two given equations, so we work on that:
From 1st equation, h = V/(4r^2).
Substitute that into the 2nd equation,
r /[H V/(4r^2)] = R/H.
Cross multiply,
rH = R[H V/(4r^2)]
rH = RH VR/(4r^2)
Isolate V,
VR/(4r^2) = RH rH
V = (RH rH) /[R/(4r^2)]
V = (4r^2)(RH rH) /R
V = (4Hr^2)(Rr) /R answer.
Or, V = (4Hr^2)(1 r/R)

3) Solve for
S= (a^n) 1 /i
a= __________
The idea here is to just isolate a.
iS = a^n 1
a^n = iS +1
Take the log of both sides,
n*log(a) = log(iS +1)
n = log(is +1) / log(a) answer.

4.) Solve for n:
P= R[(1x^n)/i]
n=___________
Just isolate n:
P/R = (1 x^(n) / i
iP/R = 1 x^(n)
x^(n) = 1 iP/R
Take the log of both sides,
(n)log(x) = log(1 iP/R)
n = log(1 iP/R) / log(x)
n = log(1 iP/R) / log(x) answer.
Can also be:
n = log[(R iP)/R] / log(x)
n = [log(R iP) +log(R)] / log(x)
n = [log(R) log(R iP)] / log(x) answer also.