# Thread: algebra refresher problems

1. ## algebra refresher problems

1. $\displaystyle ^3\sqrt{x}=4$
2. $\displaystyle 2x^2-17=-8$
3.$\displaystyle -7*2^x=-224$
4. $\displaystyle (\frac{2}{3})^x=\frac{3}{2}$
5. $\displaystyle 3^x=\frac{1}{9}$

I've picked out these problems in relation to one question. What do I do with all these indexes and powers? I totally forgot.

Thanks.

2. problem 2

$\displaystyle 2x^2 -17 = -8$

$\displaystyle 2x^2 = 9$

$\displaystyle x^2 = \frac{9}{2}$

$\displaystyle x =^+_- \sqrt{\frac{9}{2}}$

3. problem 5

$\displaystyle 3^x = \frac{1}{9}$

$\displaystyle 3^x = \frac{1}{3^2}$

$\displaystyle 3^x = 3^{-2}$

$\displaystyle x = -2$

4. problem 1

$\displaystyle ^3\sqrt{x}=4$

$\displaystyle x= 4^3$

$\displaystyle x = 64$

5. Problem 3

$\displaystyle (-7)(2^x) = -224$

$\displaystyle 2^x = 32$

$\displaystyle 2^x = 2^5$

$\displaystyle x =5$

problem 4

$\displaystyle \bigg(\frac{2}{3}\bigg)^x = \bigg(\frac{3}{2}\bigg)$

$\displaystyle \bigg(\frac{2}{3}\bigg)^x = \bigg(\frac{2}{3}\bigg)^{-1}$

$\displaystyle x = -1$