- #1

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## Homework Statement

1. (3*square root of 2)^square root of 2 How do I simplify this and other exponential expressions?

2. 3^(power of 2*x)-1=3^(power of x) + 2 How do i solve this?

## The Attempt at a Solution

I don't know how to attempt it...

- Thread starter amd123
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- #1

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1. (3*square root of 2)^square root of 2 How do I simplify this and other exponential expressions?

2. 3^(power of 2*x)-1=3^(power of x) + 2 How do i solve this?

I don't know how to attempt it...

- #2

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1. [tex](3\sqrt 2)^{\sqrt 2}[/tex]

and

2. [tex](3)^{2x}-1=3^{x}+2[/tex]

Number 1 looks like a calculator problem.

Number 2 looks like a quadratic if you let [tex]y=3^x[/tex]

I didn't work it out and I don't know exactly what your requirements are

- #3

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- #4

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Please state the original problem. Is my interpretationof the equations accurate?

- #5

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yes they are correct

- #6

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I would set that equal to something like that from the first part. GAhh its hard, MATH IS HARD!

- #7

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- #8

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ahhh i found the problem http://img365.imageshack.us/img365/1574/48510718lv6.jpg [Broken]

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- #9

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Not completely sure if I understand the problems.

1) [tex](3^{\sqrt2})^\sqrt2[/tex]

[tex](x^m)(x^n) = x^{m*n}[/tex]

so,

[tex]3^{\sqrt2*\sqrt2}[/tex]

Now just simplify it a little more.

2) [tex]3^{3x} - 1 = 3^{x+2}[/tex]

Possibly use ln to solve.

[tex]ln(3^{3x} - 1) = ln(3^{x+2})[/tex]

[tex]2xln3-(x+2)ln3 = ln1[/tex]

[tex]ln3(2x-(x+2))=ln1[/tex]

[tex]x = \frac{ln1}{ln3}+2[/tex]

1) [tex](3^{\sqrt2})^\sqrt2[/tex]

[tex](x^m)(x^n) = x^{m*n}[/tex]

so,

[tex]3^{\sqrt2*\sqrt2}[/tex]

Now just simplify it a little more.

2) [tex]3^{3x} - 1 = 3^{x+2}[/tex]

Possibly use ln to solve.

[tex]ln(3^{3x} - 1) = ln(3^{x+2})[/tex]

[tex]2xln3-(x+2)ln3 = ln1[/tex]

[tex]ln3(2x-(x+2))=ln1[/tex]

[tex]x = \frac{ln1}{ln3}+2[/tex]

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- #10

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those last equations are not correct,

[tex]ln(x+y) \not= lnx+lny[/tex]

[tex]ln(x+y) \not= lnx+lny[/tex]

- #11

HallsofIvy

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There is no "problem" there, just an expression and an equation with no indication what you are to do with them. Furthermore, it makes no sense to say "make them equal"- an expression cannot be equal to an equation.ahhh i found the problem http://img365.imageshack.us/img365/1574/48510718lv6.jpg [Broken]

[tex](3^\sqrt{2})^\sqrt{2}[/tex]

can be simplified a lot by using the "law of exponents": (a

For

[tex]3^{3x} - 1 = 3^{x+2}[/tex]

write it as

[tex](3^x)^3- 1= (3^x)(3^2)[/itex]

and let u= 3

[tex]u^3- 1= 9u[/itex]

Unfortunately, that clearly does not have any rational number solutions.

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- #12

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- #13

tiny-tim

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Your original question was (3*square root of 2)^square root of 2.

But your .jpg looks like (3^square root of 2)^square root of 2.

Which is it?

- #14

HallsofIvy

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Consider [itex]3^\sqrt{2}[/itex].

We haven't determined

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