Solve Natural Log Problem: 3^x when 2^x=3

[lLovePhysics]

Problem: If $$2^{x}=3$$, what does 3^x equal?

I solved this by taking the $$log_{2}$$ of both sides. However, the book's solution involves taking the natural log of both sides. Can someone show me why that would work? I don't get what the x power changes to a x term. The natural log is base 10 right?

 

[Kurdt]

What exactly are you asking? If 2³, what... means nothing.

Do you mean:

$2^3 = x$ then what does $3^x$ equal?

 

[lLovePhysics]

What exactly are you asking? If 2³, what... means nothing.

Do you mean:

$2^3 = x$ then what does $3^x$ equal?

My bad... I've edited the problem

 

[lLovePhysics]

Also, someone please answer this question. How come: $$\sqrt{(-2)^{2}}=2$$??

This is very mind boogling because (-2)^2=4 and the square root of 4 is 2... but, why is it wrong to simplify just by canceling out the power of 2 with the square root? I've always done it that way... =/

 

[Kurdt]

Ok no probs.

Because exponentials are defined in terms of the exponential function (hence the name) and the natural logarithm is the inverse to the exponential function, it is easy to write a logarithm to the base of any number in terms of natural logs. For example, we define $\log_a (x)=y$ to be the inverse of $x= a^y$. But from our knowledge of exponentials we can say:

$$x=a^y=e^{y\ln (a)} \rightarrow \ln(x)=y\ln(a) \rightarrow y=\frac{\ln (x)}{\ln (a)}$$

 

[Kummer]

Also, someone please answer this question. How come: $$\sqrt{(-2)^{2}}=2$$??

The mistake that many people make (especially those in physics, because math majors look out for such mistakes) is that $$\sqrt{x^2}=x$$. That is wrong. For example, if $$x=(-2)^2$$ then $$\sqrt{(-2)^2}=2$$.

The correct version states that,
$$\sqrt{x^2}=|x|$$.

 

[lLovePhysics]

Ok no probs.

Because exponentials are defined in terms of the exponential function (hence the name) and the natural logarithm is the inverse to the exponential function, it is easy to write a logarithm to the base of any number in terms of natural logs. For example, we define $\log_a (x)=y$ to be the inverse of $x= a^y$. But from our knowledge of exponentials we can say:

$$x=a^y=e^{y\ln (a)} \rightarrow \ln(x)=y\ln(a) \rightarrow y=\frac{\ln (x)}{\ln (a)}$$

So $$ln$$ is basically equal to $$log_{e}$$ right? What is $$e^{log_{e}}$$??

 

[lLovePhysics]

The mistake that many people make (especially those in physics, because math majors look out for such mistakes) is that $$\sqrt{x^2}=x$$. That is wrong. For example, if $$x=(-2)^2$$ then $$\sqrt{(-2)^2}=2$$.

The correct version states that,
$$\sqrt{x^2}=|x|$$.

Is there any reason why you can't just take cancel the sqrt. and power of 2? It is different for cube roots and cube powers right?

So the 4th root of (-4)^4 is equal to 4 right?

 

[Kurdt]

So $$ln$$ is basically equal to $$log_{e}$$ right? What is $$e^{log_{e}}$$??

Yes, ln is a shorthand for natural log or log_e. The natural logarithm is the inverse function of the exponential function and thus:

$$e^{\ln a} = a$$

 

[Kurdt]

Sorry, I didn't see in your original post that you asked if natural log was log to the base ten. Natural log is log to the base e.

 

[Mr. Mathmatics]

Is there any reason why you can't just take cancel the sqrt. and power of 2? It is different for cube roots and cube powers right?

So the 4th root of (-4)^4 is equal to 4 right?

What Kummer said was correct about the absolute value of two. You can't just cancel it out, you may cancel it out though if it is (square root of x)^2 as long as the radical symbol is inside the parentheses then you may cancel out.

 

[HallsofIvy]

Getting back to the first question:
If 2^x= 3, then log(2^x)= x log(2)= log(3) so
[tex]x= \frac{log(3)}{log(2)}[/itex]<br /> <br /> Now, 3^x= 3^{log(3)/log(2)}.<br /> <br /> That isn't going to be any simple number (it's about 5.7).[/tex]

 

[VietDao29]

Is there any reason why you can't just take cancel the sqrt. and power of 2? It is different for cube roots and cube powers right?

So the 4th root of (-4)^4 is equal to 4 right?

Yup, exactly. Well, since y = x³ is a 1-to-1 function, i.e, for every $$x_0 \neq x_1$$, we'll obtain: $$y_0 = x_0 ^ 3 \neq y_1 = x_1 ^ 3$$, so given one y, we can find only one x such that y = x³.

But y = ² is a different thing. 2 numbers a, and -a, are different (for a <> 0), but when squared, they'll become: (a²) = a², and (-a)² = (-1)² a² = a², so, we have: $$(a) ^ 2 = (-a) ^ 2 = a ^ 2$$. So, for any y (not 0), there'll be 2 different x's, one positive, one negative, such that, when squared, they'll give y.

Say, y = 4, then 2 x's are 2, and -2.

Square root function will only return the non-negative value.

Say, we have: y = 4, so sqrt(y) = 2; k = 16 ~~> sqrt(k) = 4.

So, in general, we have: $$\sqrt{a ^ 2} = |a|$$

To take the negative value, we can assign the minus sign in front of it, like this:

$$t = 36 \Rightarrow - \sqrt{t} = -6$$

-------------------------

And it's the same for even-th root, like: $$\sqrt[4]{x ^ 4} = |x| , \ \sqrt[6]{x ^ 6} = |x|, ...$$, blah, blah, blah, so on. :)

Hope that's clear. :)