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## Homework Statement

simplify m^(9*√5)/m^(√5)

## The Attempt at a Solution

would that equal m^9 or m^(8*√5)

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- Thread starter amd123
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- #1

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simplify m^(9*√5)/m^(√5)

would that equal m^9 or m^(8*√5)

- #2

symbolipoint

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- #3

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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

- #4

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please anyone?

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http://en.wikipedia.org/wiki/List_of_logarithmic_identities

- #9

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

- #11

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

No problem.

- #12

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- #13

symbolipoint

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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.

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- #15

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no i said

if i raise 1/6 to -2 then it equals 36

if i raise 1/6 to -2 then it equals 36

- #16

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

- #17

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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).

- #18

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

- #19

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

- #20

HallsofIvy

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You don't need log identities, just the fact that the exponential functions (a

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.

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