Question

^3√(2^6)^x
I know that the answer is 4^x but what are the steps to solve?

Answer 1

\(\LARGE \displaystyle \left(\sqrt[3]{2^6}\right)^x\) like this?

Answer 2

parentheses around only the 2^6

Answer 3

\(\LARGE \displaystyle\sqrt[3]{ \left(2^6\right)^x}\)

Answer 4

yes like that

Answer 5

well, this is what you can do:

\(\LARGE \displaystyle\sqrt[3]{ \left(2^6\right)^x}=\left(\sqrt[3]{ 2^6}\right)^x\)

Answer 6

then, deal with what is inside the parenthesis alone, and if the part inside the parenthesis is a 4, then the expression is 4\(^x\).

Answer 7

more percisely.

\(\LARGE \displaystyle\sqrt[3]{ \left(2^6\right)^x}=\)

\(\LARGE \displaystyle \left(\sqrt[3]{ 2^6}\right)^x=\)

\(\LARGE \displaystyle \left(2^{6/3}\right)^x=\)

\(\LARGE \displaystyle \left(2^{2}\right)^x=\)

\(\LARGE \displaystyle \left(4\right)^x.\)

Answer 8

if you have a question about the validity of the steps (i.e. what propery did I use for a particular step, or why doing something is valid and works, etc)

or other questions as wel...

Answer 9

Ok I understand it now... I just wasn't sure about about what to do with the 6 and the 3 but it makes sense now that I see it. Thank you so much!

Answer 10

Anytime.....

But, what I did in the first step is not alwyas valid, although is in this case.

Answer 11

For example, if I say:

\(\LARGE \sqrt{x^2}=\left( \sqrt{x} \right)^2\)

then I am faulty

Answer 12

Why? because the left side is the absolute value of x (i.e. |x| )
And it is continous for all values of x over the interval (-∞,+∞)

And the right side is a line y=x for all values of x, where x≥0 (but undefined for x<0)

Answer 13

But, you can always use the following property:

\(\LARGE \sqrt[a]{x^b}=x^{b/a}\)

Answer 14

that is what I used to get from line 2 to line 3