# Log Problem

1. Aug 2, 2007

### lLovePhysics

1. The problem statement, all variables and given/known data
Solve for x: log(x+5)= log x + log 5.

2. Relevant equations
log a + log b = log (ab)

3. The attempt at a solution

log (x+5)= log (5x)

x+5= 5x

4x=5

x=5/4

However, my question is: When you make the logs to the power of 10, to eliminate them, do you always need to do it to BOTH sides? Does that rule imply for any log? For example, if you want to log a number, do you need to do it to both sides? Do these kind of procedures all work the same as "Whatever you do to one side of the equation, you must do to the other?"

2. Aug 2, 2007

### lLovePhysics

Another problem that I've done:

Problem: Evaluate $$log_{27}\sqrt{54}-log_{27}\sqrt{6}$$

Solved by: $$log_{27}\sqrt{9}=log_{27}3=x$$

$$27^{x}=3$$

$$3^{3x}=3^{1}$$

Now, are you supposed to log both sides? to get 3x=1? Is that the correct way to do it?

3. Aug 2, 2007

### olgranpappy

yes, if you want the equation to remain true.

Basically, yes...

If I have an equation like
$$A=B$$
this is saying that A is the same thing as B--they are both the same number (let's say they are both equal to 4).

Let's say that I now do something to A, for example, take the positive square-root of it. So now I have a different number sqrt(A) which is typically different from A, right? Yes. 4 is different from 2. But that means sqrt(A) is different from B since I already know that A and B are the same and both are thus different from sqrt(A).

But if I take the number B (which is the same as A) and do the same thing to it, for example, take the positive square-root, then I have a number sqrt(B) which is not the same as A or B, but is the same as sqrt(A).

So, yes, in general, if I put both sides of an equation through the exact same meat grinder--any type of meat grinder whatsoever--I still get an equation out the other end. In symbols:

$$A=B$$
Implies
$$f(A)=f(B)$$
where f is any function whatsoever.

Cheers.

4. Aug 2, 2007

### olgranpappy

yep. here's another way to see that you are correct using the fact that
$$\log_{27}(27)=1$$ and the fact that 3^3=27, and the fact that (maybe you haven't seen this yet, but it's true) $$\log(a^b)=b \log(a)$$:

Since 3^3=27 we also have 3=27^(1/3) thus

$$\log_{27}(3)=\log_{27}(27^{(1/3)})=(1/3)\log_{27}(27)=(1/3)$$

Last edited: Aug 2, 2007
5. Aug 2, 2007

### olgranpappy

P.S. in tex if you use "\log" instead of "log" you will get a log symbol that looks nicer.

6. Aug 2, 2007

### lLovePhysics

Wow, thanks for all your help! =] Let me try $$\log_{27}3$$

7. Aug 2, 2007

looks nice.