Ok hopefully last time I will bug you guys about exponents . Simplify, leave without radicals and negative exponents. (sqrt(m+2))(2+m)^3/2 and (sqrt(e^(2x)) Any help is much appreciated~
Follow Math Help Forum on Facebook and Google+
Originally Posted by JonathanEyoon Ok hopefully last time I will bug you guys about exponents . Simplify, leave without radicals and negative exponents. (sqrt(m+2))(2+m)^3/2 and (sqrt(e^(2x)) Any help is much appreciated~ Try using these rules: $\displaystyle \sqrt{x} = x^{1/2}$ and $\displaystyle a^n \cdot a^m = a^{n + m}$ and $\displaystyle (a^n)^m = a^{n \cdot m}$ -Dan
Originally Posted by JonathanEyoon Simplify, leave without radicals and negative exponents. A) (sqrt(m+2))(2+m)^3/2 and B) (sqrt(e^(2x)) Hello, to A): $\displaystyle \sqrt{m+2} \cdot (2+m)^{\frac32} = (m+2)^{\frac12}\cdot (m+2)^{\frac32} = (m+2)^{\frac12 + \frac32} = (m+2)^2$ to B): $\displaystyle \sqrt{e^{2x}} = \sqrt{\left(e^x \right)^2} = e^x$
Originally Posted by topsquark Try using these rules: $\displaystyle \sqrt{x} = x^{1/2}$ and $\displaystyle a^n \cdot a^m = a^{n + m}$ and $\displaystyle (a^n)^m = a^{n \cdot m}$ -Dan Ok using the rules you gave me, I think I figured out the first one (sqrt(m+2))(2+m)^3/2 is the same as ((m+2)^(1/2))((m+2)^(3/2)) = (m+2)^2. If this is correct, from here should I foil? For the second one... Can I have a little help starting off
Originally Posted by earboth Hello, to A): $\displaystyle \sqrt{m+2} \cdot (2+m)^{\frac32} = (m+2)^{\frac12}\cdot (m+2)^{\frac32} = (m+2)^{\frac12 + \frac32} = (m+2)^2$ to B): $\displaystyle \sqrt{e^{2x}} = \sqrt{\left(e^x \right)^2} = e^x$ Hi thanks for the answer!! But I still don't understand how you got the answer for part B . Could you walk me through it?
Originally Posted by JonathanEyoon (sqrt(e^(2x)) $\displaystyle \sqrt{e^{2x}} = \left ( e^{2x} \right ) ^{1/2}$ Does this help? -Dan
Originally Posted by JonathanEyoon Ok using the rules you gave me, I think I figured out the first one (sqrt(m+2))(2+m)^3/2 is the same as ((m+2)^(1/2))((m+2)^(3/2)) = (m+2)^2. If this is correct, from here should I foil? Check with your teacher, but this form is probably fine. -Dan
Originally Posted by topsquark $\displaystyle \sqrt{e^{2x}} = \left ( e^{2x} \right ) ^{1/2}$ Does this help? -Dan Thanks!!! That helped quite a bit!
Hey final one!!.... I think haha [35(2b+1)^(3) / 7(2b+1)^(-1)]^2 I factored out a 7 and did some other stuff leaving [5(2b+1)^(4)]^2 From here I'm stuck
Originally Posted by JonathanEyoon [35(2b+1)^(3) / 7(2b+1)^(-1)]^2 I factored out a 7 and did some other stuff leaving [5(2b+1)^(4)]^2 From here I'm stuck Hello, your result is correct. Now use : $\displaystyle (a \cdot b^n)^m = a^m \cdot b^{nm}$ For confirmation only: $\displaystyle 25 \cdot (2b+1)^8$
Originally Posted by earboth Hello, your result is correct. Now use : $\displaystyle (a \cdot b^n)^m = a^m \cdot b^{nm}$ For confirmation only: $\displaystyle 25 \cdot (2b+1)^8$ From 25(2b+1)^8 Is that the simpliest form?
Originally Posted by JonathanEyoon From 25(2b+1)^8 Is that the simpliest form? Hello, I don't know a way to simplify this term, so this result must be sufficient. In short: Yes.
View Tag Cloud