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what is missing in the calculation below? i have been stuck in this question for hours, pls help. when $x=-1$, the first one gives $2$ but the 2nd one does not meet the logarithm condition. I am putting the power $2$ out in the 2nd one for some cases that it is canceled out like $0.5\cdot\log_3(-3)^2 = 0.5\cdot2\cdot\log_3(-3)$.

Is it incorrect since $-1$ from $-3$ are not squared? is there a rule that $-1$ must be powered in similar cases like this?

$$\log_3(x-2)^2=2\log_3(x-2)$$

Condition (left): $(x-2)^2 > 0$, i.e. $x\ne 2$

Condition (right): $x-2 > 0$, i.e. $x > 2$

If $x=-1$:

left: $\log_3(-3)^2=\log_39=2$

right: $2\log_3(-3)$ is undefined...

Community
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2 Answers2

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The identity:$$\log_3(x-2)^2=2\log_3(x-2)$$ holds for $x-2 > 0 \implies x>2$. Now since $-1<2$, the above is invalid for $x=-1$

Dylan Zammit
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  • $\log_3(-1-2)^2=\log(-3)^2\neq 2\log(-3)$. Yes the value for the first one is =2, and it makes sense that the second one does not give the same answer...because as I have show they are not equal, it's not even defined at $x=-1$ for that matter. – Dylan Zammit Mar 23 '18 at 14:16
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Let's call your functions $f(x)=\log_3(x-2)^2$ and $g(x)=2\log_3(x - 2)$. Even though we have the identity $\log_ab^c=c\log_ab$, notice that $f$ and $g$ are not equal functions. The domain of real $f$ is $\mathbb R\backslash\{2\}$ whereas the domain of $g$ is $\mathbb R^{>2}$. (They are only equal over the intersection of their domains.)

In your example, you are correct when you say that $f(-1)=2$ and $g(-1)$ is undefined. Although this looks like it contradicts the identity, it doesn't really: the identity simply implicitly assumes that both functions are defined.

Richard Ambler
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