Log clarification

[Baron]

Solve

log x^2 = 10

10^10 = x^2
10 000 000 000 = x^2
x = ± sqrt(10 000 000 000)
x = +100 000 or -100 000

Are both answers correct or is -100 000 an extraneous root?

P.S ... I realized I could have solved for x by bringing the exponent to the front of the log and simplifying but that would not have gotten two answers.

 

[tkhunny]

F.S. You could not have obtained both by moving the 2 outside the logarithm.

Let's think about the Domain of the logarithm.

f(x) = log(x) -- Domain: \(\displaystyle (0,\infty)\)

h(x) = 2*log(x) -- Domain: \(\displaystyle (0,\infty)\)

h(x) = log(x^{2}) -- Domain: \(\displaystyle (-\infty,0) \;OR\; (0,\infty)\)

We ask, then, can you REALLY solve by moving the 2 out?

 

[Baron]

We ask, then, can you REALLY solve by moving the 2 out?

It depends on the order of operations. If x is squared before being multiplied by -1, then -100 000 is an extraneous root.
But if x is multiplied by -1 before being squared, there are two answers.

I was just wondering because log x^2 = 2log x ... but the answers when solving are different.

I think there are two answers for the question though. Am I correct?

 

[tkhunny]

You missed the point. It has nothing to do with the order of operations.

log(x^{2}) = 2\cdot log(x) ONLY for x > 0.

When you take down the '2', you change the Domain. You have discarded all the previously valid negative values.

 

[Baron]

Since the domains of y=log x^2 and y = 2log x are different, so the equations should not be equal to each other, is the power law (bringing the exponent of a logarithm to the front) only true for some values of x? (only the positive numbers?)

 

[tkhunny]

Now we're talking.

Check your textbook carefully. When it states, log(a^b) = b*log(a), you should be able to find comments about 'a' and 'b'. In particular, a > 0.

This is a common problem when wandering through algebra. Those little things you thought were a waste of time, like the Domain, eventually will haunt you. In this case, you lost a root when applied the well-known principle, log(a^b) = b*log(a). For b = 2 and a Real and not zero, we know a^b is Positive. We just don't know that about 'a' by itself.

Excellent work finding both solutions to the ORIGINL problem. When using your best algebra, ALWAYS keep your mind and eyes open for Domain considerations. As we see here, it is NOT a small thing.