^=exponent

X=multiply

x=variable

4^(-2)

4^(-3)-4^(0)

Solve for x:

3^(2x+1-x+2)=3^(x2-x)

Anyone know how to solve these??? :confused:

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- Sep 15th 2005, 02:46 PMAnti-Math:(---Rational Exponents---:(PLZ HELP
**^=exponent**

**X=multiply**

**x=variable**

__4^(-2)__

4^(-3)-4^(0)

Solve for x:

3^(2x+1-x+2)=3^(x2-x)

Anyone know how to solve these??? :confused: - Sep 15th 2005, 06:27 PMMathGuru
Remember that x^(-1) = 1/x

Also remember that x^0 = 1

so 4^(-3) = 1/(4^3)

And

__4^(-2)__

4^(-3)-4^(0)

can be rewritten as

__1/16__

1/64-1

can you take it from there? - Sep 15th 2005, 06:29 PMMathGuru
**3^(2x+1-x+2)=3^(x2-x)**take log_3 of both sides to render the equation

2x+1-x+2 = x2-x, did you mean x^2 instead of x2 anyway you can take it from here I'm sure. - Sep 15th 2005, 06:32 PMMathGuruLearn Latex, you will love it
BTW we usually use * for multiply to avoid confusion with the variable x

you can also learn latex and write stuff like this $\displaystyle \frac{1}{3 \cdot x^{-2}}$

click on it to find out what I did. - Sep 16th 2005, 01:46 PMAnti-Math
Thnx