Find the root of this equation:
log(x-2)=1

Question

Find the root of this equation: 
log(x-2)=1

parthkohli · Answer

Base $10$? You will have:$$\log(x - 2) = 1\iff 10^1 = x - 2$$I'm sure it is easy enough for you to follow.

anonymous · Answer

Here's an example of the method my teacher would like for me to follow

anonymous · Answer

x=12 ?

parthkohli · Answer

Have you heard about the exponential form to logarithmic form? Basically, you can show this:$$\log(x - 2) = 1 \tag{1}$$$\textbf{AXIOM:}$ $\log_a b = c \iff a^c = b$.$$\therefore \log (x - 2) = 1 \iff 10^1 = x-2$$

parthkohli · Answer

Yes, $x = 12$.

anonymous · Answer

Oh yes thank you:) I was just confused because I usually have question involving the log of both sides

parthkohli · Answer

Oh, good enough. :-)

anonymous · Answer

@ParthKohli  how would i find the restrictions ?

parthkohli · Answer

Restrictions? For the function $f(x) = \log(x - 2)$?

anonymous · Answer

Yes

parthkohli · Answer

$\log(x)$ is restricted to non-negative values for $x$.

parthkohli · Answer

So when $x - 2 \ge 0$, the function is defined.

anonymous · Answer

look in the attachment i posted earlier

anonymous · Answer

Kind of like that

parthkohli · Answer

Actually the restrictions are $x < 0$ instead of $x>0$.

agent0smith · Answer

@parthkohli "So when x−2≥0, the function is defined."

I think that should be when x-2>0, as log x is only defined for x>0.

parthkohli · Answer

@agent0smith Hmm, that's a very active argument. But,$$\log(0) = -\infty$$