Confusion in solutions of logarithm.

Question

An expression $log_2 x^2=2$ can be written as $2log_2 x=2$ but leads to loss of a root. I am having difficulty is recognizing the expressions in which above property is applicable. Another expression $log_2^3 x=log_2 x^3$ can be written as $log_2^3 x=3log_2 x$ and doesn't leads to any loss. My teacher has finished logarithm and it is very difficult to ask him as he rarely replies to queries(also the teaching is online).

I thought that it would be related to the power of $x$ if it is either odd or even i.e. we apply property if the power is odd but not when it is even. And also if the base is a constant or variable i.e. property applicable to logs with variable bases and not in constant bases.

I want to know how to recognize in questions that this property doesn't lead to any loss. Kindly help.

Answer 1

You seem to be assuming the logarithm power law holds for any real numbers. This isn't generally true. What we can say is that $$\log_a{(x^n)}=n\log_a{(x)}$$ holds for all $a,x\gt0$ (with $a\ne1$) and $n\in\mathbb{R}$. But you can also instead use $|x|\ge0$ to somewhat extend this identity to the entire non-zero reals. Namely $$\log_a{(|x|^n)}=n\log_a{(|x|)}$$ holds for all $x\in\mathbb{R}\setminus\{0\}$. For even $n\in\mathbb{Z}$ this becomes $$\log_a{(x^n)}=n\log_a{(|x|)}$$

Answer 2

hint

For any $ X>0$, we know that

$$\log_2(X^2)=2\log_2(X)$$

For any $ x\ne 0$, we have

$$\log_2(x^2)=\log_2(|x|^2)=2\log_2(|x|)$$

For $ x^3 $, we have $ x^3=|x|^3$, only if $ x>0$.

Answer 3

Simply put:

We can NOT write the expression $\log_2 x^2$ as $2\log_2 x$. And if we are given an equation $\log_2^3 x = \log_2 x^3$ that is a specific problem to solve and, in general $\log_2^3 m \ne \log_2 m^3$.

Details

An expression $\log_b m$ will (if we are working with real numbers) not make sense unless $b >0$ and $b\ne 1$ and if $m > 0$. We have an identity law that if $m> 0$ then $\log_b m^k = k \log_b m$. That's an identity and it does require that $b, m > 0$ and that $b\ne 1$.

Now when we express $\log_2 x^2 = something$ this is a equation we need to solve. We off the bat do not no anything about $x$. To solve it we note:

For the expression to be valid we know $x^2 > 0$ so $x \ne 0$. We can NOT say $\log_2 x^2 = 2\log_2 x$ because we do not know if $x > 0$ and if $x < 0$ hen $2\log_2 x$ is not a valid statement at all.

But we do know that $x^2 = |x|^2$ and $|x| \ge 0$ and as $x\ne 0$ then $|x|>0$ and so we CAN say $\log_2 x^2 = \log_2 |x|^2 = 2\log_2 |x|$.

(And then we can solve $2\log_2 |x| = something$ so $\log_2 |x| = \frac {something}2$ so $|x| = 2^{\frac {something}2}$ and $x=\pm 2^{\frac{something}2}$.)

But if we have an equation $\log_2 x^k=something$ where $k$ is odd is a slightly different story.

If $\log_2 x^3 = something$ we know that $x^3 > 0$ and therefore that $x > 0$. So in that case we can write $\log_2 x^k = k\log_2 x$ because we DO know that $x > 0$.

So $\log_2 x^3 = something$ can be solved $3\log_2 x = something$ and $\log_2 x = \frac {something}3$ and $x = 2^{\frac {something}3}$.

We should note that if we have the equation $\log_2 (-x^3)=something$ or $\log_2 (-x)^3 = something$ then we know that $(-x^3) >0$ and $x^3 < 0$ (or $(-x)^3 > 0$ so $-x > 0$ and $x < 0$) and we can still continue.

If we have the equation $\log_2^3 x =\log_2 x^3$ then this is an unusual equation and is NOT true in general. $\log_2^3 4 \ne \log_2 4^3$ and in general $\log_2^3 m \ne \log_2 m^3$. So if we have $\log_2^3 x = \log_2 x^3$ this is a problem where we are expected to solve for $x$.

For the expression $\log_2^3 x$ to make sense we must have $x >0$. ANd for that expression $\log_2 x^3$ to make sense we must have $x^3 >0$. So in either even we know $x>0$.

Since we KNOW $x > 0$ we can say that $\log_2 x^3 =3\log_2 x$ and we have $\log_2^3 x = 3\log_2 x$. If we let $m = \log_2 x$ we must solve $m^3 = 3m$. (But we must have $x > 0$)

The solutions to that are $m = 0$ or if $m \ne 0$ then $m^2 = 3$ so $m = \pm \sqrt 3$.

So we have $\log_2 x = 0, \pm \sqrt 3$. $\log_2 x = 0$ means $x=1$ and $\log_2 x = \pm \sqrt 3$ means $x = 2^{\pm \sqrt 3}$. All of these values are positive so they are all valid.

Note if we had $\log_3^2 x = \log_3x^2$ we'd solve it by noting, for $\log_3^2 x$ to make sense $x > 0$. And for $\log_3x^2$ to make sense we must have $x^2 > 0$. But $x > 0$ so that is redundant.

Then we can say, because we know $x> 0$ that $\log_3 x^2 =2\log_3 x$ and that $\log_3 x = \log_3^2 x$. But $\log_3^2 x = (\log_3 x)^2\ge 0$ because it is a square.

So if we let $\log_3 x = m$ we know $m \ge 0$ and we need to solve $2m = m^2; m \ge 0$. So $m=0$ or $m=2$. So we have $\log_3 x = 0$ or $\log_3 x =2$. So $x = 3^0=1$ or $x = 3^2 = 9$.