How do I evaluate $\log_3(x^2 − 36) − \log_3(x − 6)?$

Question

The questions ask to combine the expression using the laws of logarithms. $$\log_3(x^2 − 36) − \log_3(x − 6)$$

I got $$\frac{\log_3(x^2-36)}{\log_3(x-6) } $$ but my book says that is not correct.

Did you remember that $\log\frac ab = \log a - \log b$? Your expression has the latter form. — MJD, Mar 31 '14 at 03:38
Well if I make the expression a quotient it still is incorrect according to my book. — Frustrated with Math, Mar 31 '14 at 03:51
Perhaps you should say in your question what quotient you got. Otherwise nobody will know how to correct your error. Or perhaps there is no error and your book is mistaken. But we will never know if you don't tell us. — MJD, Mar 31 '14 at 03:52
$ \log a - \log b $ is equal to $\log\frac ab$. It is not equal to $\frac{\log a}{\log b}$. — MJD, Mar 31 '14 at 03:55
It might also be worth pointing out that $\log a - \log b$ can't possibly be equal to $\frac{\log a}{\log b}$, because subtraction is not the same thing as division. — MJD, Mar 31 '14 at 04:26
I think you need to say more than just subtraction is not division. For example $4/2=4-2=2.$ — R R, Mar 31 '14 at 04:44
@RR Implicit universal quantifier. Though it would have been good to make that quantifier explicit. — David Richerby, Mar 31 '14 at 07:31

score 8 · Answer 1 · edited Mar 31 '14 at 07:51

8

Recall that $$\log a-\log b=\log\left(\dfrac{a}{b}\right)$$

Then using this principle, $$\log_3(x^2 − 36) − \log_3(x − 6)=\log_3\left(\dfrac{x^2-36}{x-6}\right)=\underbrace{\log_3\left(\dfrac{(x+6)(x-6)}{x-6}\right)}_{\text{Here the } (x-6)\text{'s cancel}}=\log_3(x+6)$$

edited Mar 31 '14 at 07:51

SQB

2,094

answered Mar 31 '14 at 04:00

Sujaan Kunalan

10,944

2

What about when $x=6$? – SQB Mar 31 '14 at 07:51
1

That is not in domain as log(x-6) is not defined. – evil999man Mar 31 '14 at 08:03
1

@Awesome That's exactly my point, since $x=6$ is in the domain for the resulting equation. – SQB Mar 31 '14 at 09:53

score 2 · Answer 2 · edited Mar 31 '14 at 07:35

2

The rule of logarithms for division dictates that $$\log_b {m} - \log_b n=\log_b \frac{m}{n}$$ for any base $b$ and any non-zero numbers $m,n$.

Substitute: $m = x^2-36$, $n=x-6$, and $b=3$. So

$$\log_3 (x^2-36) - \log_3(x-6)=\log_3 \frac{x^2-36}{x-6}=\log_3 \frac{(x+6)(x-6)}{x-6}=\log_3 (x+6)$$

edited Mar 31 '14 at 07:35

David Richerby

2,450

answered Mar 31 '14 at 04:01

Cookie

13,532

What about when $x=6$? – SQB Mar 31 '14 at 07:52

score 2 · Answer 3 · answered Mar 31 '14 at 06:46

2

You applied the formula in a wrong way. The formula does not say that the difference of two logarithms is the quotient of the two logarithms. It says that the difference of two logarithms is the logarithm of the quotient of the two arguments !

answered Mar 31 '14 at 06:46

1

By the way, if $\log(a)-log(b)=\frac{\log(a)}{\log(b)}$, that would mean that $A-B=\frac{A}{B}$ for any numbers ! – Mar 31 '14 at 06:50

How do I evaluate $\log_3(x^2 − 36) − \log_3(x − 6)?$

3 Answers3