Need ALGEBRA 2 HELP with square roots

You are confused by your own notation. If you mean the numerator is m^(9*(5)^0.5) and the denominator is m^((5)^(.5)), then the simplification is m^(8*(5)^(.5)). This is your second alternative which you are deciding.

 

x^y/x^z = x^(y-z) Dividing two exponents is subtracting their exponents. As for the logarithm problems, you might want to look up the properties of logs and exponents and play around with them.

 

I think it's fine to leave the square roots alone, unless you want to give a non-exact value. The second answer in your original post looks right.

 

Take a look at the following for logarithm identities:
http://en.wikipedia.org/wiki/List_of_logarithmic_identities

 

I don't understand what you mean by making the base equal to 36. What I would do is use logarithms. If a = b, then ln(a) = ln(b). Then you can use the log identities to simplify and solve for x.

No problem.

 

Use the base of 6. On the leftside, you have 6^(-x). On the rightside you have (6^2)^(x+3)=6^(2(x+3))

You then simply have 6^(-x) = 6^(2(x+3))

*Understand the reasoning that lead to that.
*Finish the solution process.

 

If you use what symbolipoint wrote, you get -x = 2(x+3). Then you just solve for x.

Or if you use base 36, then 1/6 = 36^-.5, not -2.

Edit: x = -2 if you use base 6 (doing what symboli says).

 

You don't need log identities, just the fact that the exponential functions (a^x for any positive a) are "one-to-one"

If you change 36 to base 1/6, then you have 36= (1/6)^(-2) so 36^(x+3)= ((1/6)^(-2))(x+3) and so (1/6)^(-2(x+3))= (1/6)^x. Since that function is "one-to-one" (two DIFFERENT values of the exponent can't give the same value), -2(x+3)= x.

I would have been inclined to use 6 as base (just don't like fractions, I guess). Then 1/6= 6^(-1) so (1/6)^x= 6^(-x). 36= 6^2 so 36^(x+3)= 6^(2(x+3)). Now you have 6^(-x)= 6^(2(x+3)). That means -x= 2(x+3). Of course , those two equations have exactly the same solution.