Homework Help: Need ALGEBRA 2 HELP with square roots

amd123 · May 3, 2008

1. The problem statement, all variables and given/known data
simplify m^(9*√5)/m^(√5)

3. The attempt at a solution
would that equal m^9 or m^(8*√5)

symbolipoint · May 3, 2008

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.

amd123 · May 3, 2008

ahhh hold on let me write it again

m ^ (9 X √5) DIVIDED by m ^ (√5) this problem involves properties of exponents and its a multiple choice problem so i know the answer can not be this m^(8*(5)^(.5)).

Also i need help solving this:
1. 5Log 5 65
2. Log (1/3)x = -1
3. 5^((log 5 2x - log 5 (x-3)) = ln e x+4
4. 63n = 435n-4
5. 4 + 3e5x = 27

amd123 · May 3, 2008

please anyone?

bubbles · May 3, 2008

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.

amd123 · May 3, 2008

you didnt even address how to do things with square roots in them and i figured out how to do 4 and 5 :)

bubbles · May 3, 2008

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.

bubbles · May 3, 2008

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

amd123 · May 3, 2008

right now im stuck on this problem (1/6)^x = 36^(x+3) if i solve it by making the bases equal to 36 by multiplying 1/6 by -2 i get -2x but then for x i get -3 which obviously would not work, but if i use -2 as x the equation works BUT how do i get -2 from that equation????

amd123 · May 3, 2008

and thanks for your help with the first one :) really appreciate it

bubbles · May 3, 2008

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.

amd123 said: ↑

and thanks for your help with the first one :) really appreciate it

No problem.

amd123 · May 3, 2008

my alg2 teacher said inorder to solve (1/6)^x = 36^(x+3) you need to make the BASES equal to each other i can easily make 1/6 to 36 by raising it to -2 but then i have to multiply x by -2. How do the log identities work?

symbolipoint · May 3, 2008

amd123 said: ↑

right now im stuck on this problem (1/6)^x = 36^(x+3) if i solve it by making the bases equal to 36 by multiplying 1/6 by -2 i get -2x but then for x i get -3 which obviously would not work, but if i use -2 as x the equation works BUT how do i get -2 from that equation????

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.

amd123 · May 3, 2008

yes symboli but then i get x = -1 and i know that x has to be -2, before when i made both bases equal to 36 i got -3 and that makes 36^0. so how do i get the equation to get me -2 for x?

amd123 · May 3, 2008

no i said
if i raise 1/6 to -2 then it equals 36

amd123 · May 3, 2008

but doing that and what symboli says only gets me -3 or -1 not -2 :(?!?

bubbles · May 3, 2008

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).

amd123 · May 3, 2008

lol i just did it and got it this is why i should not use a pen with math :P

amd123 · May 3, 2008

now if only i knew how to do this 6^(3n) = 43^(5n-4)

HallsofIvy · May 4, 2008

amd123 said: ↑

my alg2 teacher said inorder to solve (1/6)^x = 36^(x+3) you need to make the BASES equal to each other i can easily make 1/6 to 36 by raising it to -2 but then i have to multiply x by -2. How do the log identities work?

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.