# Exponential Functions

1. 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?

3. The Attempt at a Solution
I don't know how to attempt it...

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so you have
1. $$(3\sqrt 2)^{\sqrt 2}$$
and
2. $$(3)^{2x}-1=3^{x}+2$$
Number 1 looks like a calculator problem.
Number 2 looks like a quadratic if you let $$y=3^x$$
I didn't work it out and I don't know exactly what your requirements are

there are none for 1 i have to simplify and for 2 i have to find the value of x that makes both of those equal

you have to make 1 and 2 equal?
Please state the original problem. Is my interpretationof the equations accurate?

yes they are correct

basically i have to find the value of x and by taking what the number is being multiplied to the power of so 3^x +2
I would set that equal to something like that from the first part. GAhh its hard, MATH IS HARD!

please state the problem as it is written. I'm not seeing the connection here. I want to help you out.

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

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Not completely sure if I understand the problems.

1) $$(3^{\sqrt2})^\sqrt2$$

$$(x^m)(x^n) = x^{m*n}$$

so,
$$3^{\sqrt2*\sqrt2}$$

Now just simplify it a little more.

2) $$3^{3x} - 1 = 3^{x+2}$$

Possibly use ln to solve.

$$ln(3^{3x} - 1) = ln(3^{x+2})$$

$$2xln3-(x+2)ln3 = ln1$$

$$ln3(2x-(x+2))=ln1$$

$$x = \frac{ln1}{ln3}+2$$

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those last equations are not correct,
$$ln(x+y) \not= lnx+lny$$

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

$$(3^\sqrt{2})^\sqrt{2}$$
can be simplified a lot by using the "law of exponents": (ab)c= abc. What is $\sqrt{2}\sqrt{2}$?

For
$$3^{3x} - 1 = 3^{x+2}$$
write it as
[tex](3^x)^3- 1= (3^x)(3^2)[/itex]
and let u= 3x. Then your equation becomes
[tex]u^3- 1= 9u[/itex]
Unfortunately, that clearly does not have any rational number solutions.

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hmm omg this is simple alg 2 why can't I get this? Now I can't understand logartithms :( I'm gonna have to use google to search some of this up.

tiny-tim
Homework Helper
Hi amd123!

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?

HallsofIvy
Consider $3^\sqrt{2}$. If that is rational, we are done. Of course, I don't happen to know whether it is rational or not, but that doesn't matter because if it is not rational then $(3^\sqrt{2})^\sqrt{2}$ is "an irrational number to an irrational power" and is clearly rational.