I'm stuck on such a teensy little thing which is make bases the same then proceding to solve for the exponents. Here is the equation i have to solve
(2x^2) ^ 3-2m x (1/ X) ^ 2m
how do i balance (2x^2) and (1/x) ?
I'm stuck on such a teensy little thing which is make bases the same then proceding to solve for the exponents. Here is the equation i have to solve
(2x^2) ^ 3-2m x (1/ X) ^ 2m
how do i balance (2x^2) and (1/x) ?
First, please never use 'x' for multiplication in the same equation as 'X' for a variable. Very confusing. Look very carefully at what you have written. You became confused in the middle.
Second, 1/X = X^(-1)
[QUOTE=Also sprach Zarathustra;661312]We are also![/Q
Ok so let me present it again...sorry for the confusion
(2X^(exponent) 2) ^(exponent) 3-2m * (times) (1/ X) ^ (exponent) 2m
i have to make the bases which are (2X^(exponent) 2) and (1/ X)
[QUOTE=NewMath;661319]$\displaystyle ( 2x^2)^{3-2m} (\frac{1}{ x}) ^{ 2m}=2^{3-2m}x^{2(3-2m)}(\frac{1}{ x^{ 2m}}) =2^{3-2m}x^{6-4m}\frac{1}{ x^{ 2m}}=2^{3-2m}x^{6-4m}x^{-2m}=2^{3-2m}x^{6-4m-2m}=2^{3-2m}x^{6-6m}=2^32^{-2m}x^6x^{-6m}=\frac{2^3}{2^{2m}}\frac{x^6}{x^{6m}}=\frac{8}{ 2^{2m}}(\frac{x}{x^{m}})^6 $
It's not perfectly clear what is wanted. I'll give it a try.
$\displaystyle (2x^{2})^{3-2m}\cdot \left(\frac{1}{x}\right)^{2m}\;=$
$\displaystyle 2^{3-2m}\cdot x^{6-4m}\cdot \left(x^{-1}\right)^{2m}\;=$
$\displaystyle 2^{3-2m}\cdot x^{6-4m}\cdot x^{-2m}\;=$
$\displaystyle 2^{3-2m}\cdot x^{6-4m-2m}$
What else?