- #1

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3 x √27 = 3^N

this can be written as

3 x (√3 x √3 x √3)

which makes

9√3 = 3^N

now where though

i presume n must be a fraction somthing like x/2 ? x is a number? Is this right. How do i go about finding the answer?

- Thread starter thomas49th
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- #1

- 655

- 0

3 x √27 = 3^N

this can be written as

3 x (√3 x √3 x √3)

which makes

9√3 = 3^N

now where though

i presume n must be a fraction somthing like x/2 ? x is a number? Is this right. How do i go about finding the answer?

- #2

Hurkyl

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Of only you could rewrite the equation as 3^{something} = 3^N, it would be obvious...

- #3

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huh? Sorry I dont see...was that a suggestion/hint or comment

as ive got a root it does the n have a 2 as it's denominator?

as ive got a root it does the n have a 2 as it's denominator?

Last edited:

- #4

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If you're not that clever student, take logs. Solve from there. Then see if you can see what the clever student would do.

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- #7

cristo

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Well, you're forced to be the clever student then, aren't you! Hurkyl's post was a hint. To put it more blatantly-- try writing the left hand side as 3 to the power of something. Can you write sqrt(3) as 3^x, for some x? Can 9 be expressed as 3 to some power? Finally, can you express a product of two numbers with the same base as one number?

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- #9

Gib Z

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- #10

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- #11

HallsofIvy

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Finally, if you still don't understand,

[tex]\sqrt{a}= a^{\frac{1}{2}}[/tex].

[tex]\sqrt{a}= a^{\frac{1}{2}}[/tex].

- #12

Gib Z

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Even More Finally if you still don't get it, give up.

N=2.5

N=2.5

- #13

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I dont see what that means...x->3^x is a well defined map,

I know that [tex]\sqrt{a}= a^{\frac{1}{2}}[/tex] but because I have a whole number and a surd im not sure what to do. Can someone just WRITE OUT the steps they did one by one using latex?

Thx

- #14

Gib Z

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[tex]\sqrt{27}=3\sqrt{3}[/tex]

Therefore:

[tex]9\sqrt{3}=3^n[/tex]

[tex]3^2 \cdot 3^{\frac{1}{2}} = 3^n[/tex]

Since [tex]a^m \cdot a^n = a^{m+n}[/tex], then

[tex]3^2 \cdot 3^{\frac{1}{2}} = 3^{2.5}[/tex]

[tex]3^{2.5} = 3^n[/tex]

[tex]n=2.5[/tex]

- #15

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NICE! Yeh I see now...ingenius.

Thanks for all the help

Thanks for all the help

- #16

HallsofIvy

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[tex]a^x a^y= a^{x+ y}[/tex]

and

[tex](a^x)^y= a^{xy}[/tex]

(I am not saying that Gib Z

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