Using Log to Undo an Exponent - Ex 2

[Jason76]

\(\displaystyle x^{4} = 8\)

I'm assuming the step here is to take the \(\displaystyle log\) of both sides. But the \(\displaystyle log\) of what? \(\displaystyle x\)?

 

[Jason76]

RULE: if a^p = b then a = b^(1/p)
x = 8^(1/4)

Why don't you take time to look up basic exponent rules...

In this case, you wouldn't use \(\displaystyle log\). You would just take the \(\displaystyle 4th\) root of each side.

\(\displaystyle x^{4} = 8\)

\(\displaystyle 4th\) root of \(\displaystyle x^{4} =\) 4th root of \(\displaystyle 8\) - (sorry don't know latex for roots other than the square one)

\(\displaystyle x = 2\)

 

[JeffM]

In this case, you wouldn't use \(\displaystyle log\). You would just take the \(\displaystyle 4th\) root of each side.

\(\displaystyle x^{4} = 8\)

\(\displaystyle 4th\) root of \(\displaystyle x^{4} =\) 4th root of \(\displaystyle 8\) - (sorry don't know latex for roots other than the square one)

\(\displaystyle x = 2\)

Inside "\(\displaystyle ...\)," use \sqrt[4]{x^4} to render \(\displaystyle \sqrt[4]{x^4}.\)

But the 4th root of 8 is not 2 because 2 * 2 * 2 = 8.

 

[Jason76]

EDITED

Changing to a different problem to show the mechanics of such a situation:

\(\displaystyle x^{3} = 8\)

\(\displaystyle \sqrt[3]{x^3} = \sqrt[3]{8}\)

\(\displaystyle x = 2\)

Never mind the \(\displaystyle x^{4}\) stuff cause that doesn't work out.

 

[lookagain]

In this case, you wouldn't use \(\displaystyle log\). You would just take the \(\displaystyle 4th\) root of each side.

\(\displaystyle x^{4} = 8\)

\(\displaystyle 4th\) root of \(\displaystyle x^{4} =\) 4th root of \(\displaystyle 8\) - (sorry don't know latex for roots other than the square one)

\(\displaystyle > > x = 2 < < \)

Jason76, you are missing another point. If you are solving for all real solutions, and the exponent on the variable is a nonzero even number and it is equated to a positive real number, then there will be two real solutions, same magnitude but opposite in sign.

Example:


\(\displaystyle x^4 = 16\)

\(\displaystyle x = \pm\sqrt[4]{16}\),

\(\displaystyle x = \pm\sqrt[4]{(2)^4}\)

\(\displaystyle x = \pm 2 \)

 

[JeffM]

EDITED

Changing to a different problem to show the mechanics of such a situation:

\(\displaystyle x^{3} = 8\)

\(\displaystyle \sqrt[3]{x^3} = \sqrt[3]{8}\)

\(\displaystyle x = 2\)

Never mind the \(\displaystyle x^{4}\) stuff cause that doesn't work out.

Yes your revised example is worked out correctly. However your working does not show the reasoning involved

\(\displaystyle x^3 = 8 \implies\)

\(\displaystyle x^3 = 2^3\implies\)

\(\displaystyle \sqrt[3]{x^3} = \sqrt[3]{2^3} \implies\)

\(\displaystyle x = 2.\)

 

[mmm4444bot]

Never mind the \(\displaystyle x^{4}\) stuff cause that doesn't work out.

It would work out, if you were to do it correctly.

Are you just making stuff up now? I mean, from where are these recent questions coming? Are you in a math class yet? :cool:

 

[Jason76]

It would work out, if you were to do it correctly.

Are you just making stuff up now? I mean, from where are these recent questions coming? Are you in a math class yet? :cool:

\(\displaystyle x^{4} = 8\)

\(\displaystyle x = \sqrt[4]{8}\)

This works out to something.

But \(\displaystyle x^{3} = 8\)

comes out to a natural number that more people are familar with.

\(\displaystyle x = \sqrt[3]{8}\)

\(\displaystyle x = 2\)

Because \(\displaystyle 2^{3} = 8\) and \(\displaystyle 8^{1/3} = 2\) and \(\displaystyle \log_{2}8 = 3\)

 

[mmm4444bot]

\(\displaystyle x^{4} = 8\)

This works out to something.

But \(\displaystyle x^{3} = 8\)

comes out to a natural number that more people are familar with.

Ah, you were trying to create an example involving a fourth-power with a Natural-number solution?

x^4 = 1

x^4 = 16

x^4 = 81

x^4 = 625

x^4 = 1296

x^4 = et cetera

(Think about it.)

 

[lookagain]

\(\displaystyle x^{4} = 8\)

\(\displaystyle x = \sqrt[4]{8}\) \(\displaystyle \ \ \ \ \) No, Jason76, \(\displaystyle \ x \ = \ \pm \sqrt[4]{8}.\)

This works out to something.

When taking the even root, don't forget the plus-or-minus sign.