Write $y = 1/2\log_a(x) + 1/2\log_a(y) - 3/4\log_a(z)$ as a single logarithm, and state restrictions on the variable.

Question

Write $y = \frac 12\log_a(x) + \frac 12\log_a(y) - \frac 34\log_a(z)$ as a single logarithm, and state restrictions on the variable.

Having trouble with the second part of this question, stating the variables. Got it reduced to $$\log_a\left(\frac{x^{\frac 12} \cdot y^{\frac 12}}{z^{\frac 34}}\right)$$ using the logarithm laws. However, I'm not sure how to make the denominator of the argument equal $0$. I believe $a$ must be greater than $0$, and as far as I understand the numerator can be a any value, but I'm not sure about that $z$ value and if there are any values that would make it equal to $0$ causing us to divide by $0$.

Answer 1

In the expanded formula:

• The base $a$ logarithm is a notation for $\log_a(u)=\dfrac{\ln(u)}{\ln(a)}$ so you need $a>0$ and also $\ln(a)\neq 0\iff a\neq 1$

• For all the logarithms to be defined we must have $x,y,z>0$.

In the condensed formula:

• You also need $a>0$ and $a\neq 1$ for the same reason

• Since $u\mapsto u^2$ and $u\mapsto u^4$ are positive even functions, their reciprocals cannot handle negative values, thus $x^\frac 12,y^\frac 12,z^\frac 34$ need $x,y,z\ge 0$ to be well defined.

• For the fraction to be defined you need in addition that $z^\frac 34\neq 0\iff z\neq 0$

• For the logarithm to be defined you need the numerator $>0\implies x,y>0$ (since the denominator is already $>0$ from points above).

You can check that eventually the restrictions are the same in both formulas.

Answer 2

For $\log_a M$ to be well defined we must have $a$ (which in context, I presume is not a variable but to be treated as constant) $a > 0$ and $a \ne 1$. And $M > 0$. Those are the restriction.

So to have $\frac 12 \log_a x +\frac 12 \log_a y - \frac 34 \log_a z$ we must have $x,y, z > 0$. That's a restriction from then get go.

That is equal to $\log_a x^{\frac 12} + \log_a y^{\frac 12} -\log_a z^{\frac 34}$.

To have $M^{\frac m{2k}}$ to be well defined $M$ must be non-negative. But we already had the restriction that $x,y,z > 0$ from the beginning so there is no know restriction.

ANd that's equal to $\log_a \frac {x^{\frac 12}y^{\frac 12}}{z^{\frac 34}}$ which requires $\frac {x^{\frac 12}y^{\frac 12}}{z^{\frac 34}} > 0$ (which we already had) and the $x^{\frac 34}$ which we already had. So no more restrictions need to be stated.

But note we could also do

$\log_a \frac {x^{\frac 12}y^{\frac 12}}{z^{\frac 34}} = \log_a (\frac {xy}{z^{\frac 32}})^{\frac 12}=\frac 12\log_a \frac {xy}{z^{\frac 32}}$.

As written as such we could have $x, y < 0$ and $x,y > 0$. But this is not allowed by our initial restrictions. So we need to state that $x,y$ are both positive.

I suppose we could also get $\frac 18 \log_a \frac {x^4y^4}{z^6}$ (but that's just stupid......) and we'd have to specify that $z > 0$ too... but as $z$ is to $\frac 12$ power or an odd power while $x,y$ to even power, I can't see a natural way that we'd have to specify that $z$ must be positive as it's dang near (but not completely) impossible to express the value so that $z < 0$ is an option.