# Need ALGEBRA 2 HELP with square roots

amd123

## Homework Statement

simplify m^(9*√5)/m^(√5)

## The Attempt at a Solution

would that equal m^9 or m^(8*√5)

Homework Helper
Gold Member
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
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

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

bubbles
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.

amd123
right now I am 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
and thanks for your help with the first one :) really appreciate it

bubbles
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.

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

No problem.

amd123
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?

Homework Helper
Gold Member
right now I am 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
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
no i said
if i raise 1/6 to -2 then it equals 36

amd123
but doing that and what symboli says only gets me -3 or -1 not -2 :(?!?

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

amd123
now if only i knew how to do this 6^(3n) = 43^(5n-4)