# Natural log problem

1. Jul 25, 2007

### 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?

Last edited: Jul 25, 2007
2. Jul 25, 2007

### Kurdt

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

Do you mean:

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

3. Jul 25, 2007

### lLovePhysics

My bad... I've edited the problem

4. Jul 25, 2007

### 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... =/

Last edited: Jul 25, 2007
5. Jul 25, 2007

### Kurdt

Staff Emeritus
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:

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

6. Jul 25, 2007

### Kummer

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,
$$\sqrt{x^2}=|x|$$.

7. Jul 25, 2007

### lLovePhysics

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

8. Jul 25, 2007

### lLovePhysics

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?

9. Jul 25, 2007

### Kurdt

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

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

Last edited: Jul 25, 2007
10. Jul 25, 2007

### Kurdt

Staff Emeritus
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.

11. Jul 25, 2007

### Mr. Mathmatics

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.

12. Jul 26, 2007

### HallsofIvy

Staff Emeritus
Getting back to the first question:
If 2x= 3, then log(2x)= x log(2)= log(3) so
$$x= \frac{log(3)}{log(2)}[/itex] Now, 3x= 3log(3)/log(2). That isn't going to be any simple number (it's about 5.7). 13. Jul 26, 2007 ### VietDao29 Yup, exactly. Well, since y = x3 is a 1-to-1 function, i.e, for every [tex]x_0 \neq x_1$$, 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:

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

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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. :)

Last edited: Jul 26, 2007