How to interpret "has a solution" in this exercise?

Question

Is the following question poorly written?

I suppose it is a translation since it comes from a Japanese institution. I am referring to the assumption "... has a solution". Given this information I would assume that the discriminant is equal to $0$ since the equation has "a solution" but the answer involves writing it with a "greater than or equal to" symbol.

What are your opinions on this? Am I wrong or should the translation specify that it has one or two solutions?

Answer 1

The phrase "has a solution" in your exercise means "there exists at least one" solution. It does not necessarily mean that this is the only one. See for instance the following excerpt from the Handbook of Mathematics:

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Observe that $$ \begin{align} \log_{10}(ax)\log_{10}(bx) &=\frac{\ln(ax)\ln(bx)}{(\ln 10)^2}\\ &=\frac{(\ln a+y)(\ln b+y)}{(\ln 10)^2}\\ &=\frac{y^2+(\ln a+\ln b)y+\ln a\ln b}{(\ln 10)^2},\quad y:=\ln x \end{align} $$ and hence $$ {y^2+(\ln a+\ln b)y+\ln a\ln b}=-(\ln 10)^2. $$ So the discriminant $$ \Delta = (\ln a+\ln b)^2-4(\ln a\ln b+(\ln 10)^2) =(\ln b-\ln a)^2-4(\ln 10)^2 \ge 0. $$ It follows that $$ |\ln b-\ln a|\ge 2{\ln 10}. $$ But $\ln b-\ln a=\ln\frac{b}{a}$. So $$ \ln\frac{b}{a}\geq A\quad \textrm{or }\quad \ln\frac{b}{a}\leq -A,\quad A:= 2{\ln 10}. $$ Therefore, $$ \frac{b}{a}\geq e^A\quad \textrm{or }\quad 0<\frac{b}{a}\le e^{-A}. $$

One can further simplify that $$ e^A=10^2=100,\quad e^{-A}=\frac{1}{100} $$

Answer 2

Set $A=\log_{10}a$, $B=\log_{10}b$ and $X=\log_{10}x$. Then the equation becomes $$ (A+X)(B+X)+1=0 $$ that simplifies into $X^2+(A+B)X+AB+1=0$. Since we know that a solution exists, we need $(A+B)^2-4AB-4\ge0$, which becomes $$ (B-A)^2\ge4 \tag{*} $$ Why $B-A$? Because $$ B-A=\log_{10}b-\log_{10}a=\log_{10}\frac{b}{a} $$ Condition (*) translates to $$ \log_{10}\frac{b}{a}\le -2 \qquad\text{or}\qquad \log_{10}\frac{b}{a}\ge2 $$ hence to $$ 0<\frac{b}{a}\le\frac{1}{100}\qquad\text{or}\qquad\frac{b}{a}\ge100 $$ Of course one could switch $a$ and $b$; the solution is, equivalently, $$ 0<\frac{a}{b}\le\frac{1}{100}\qquad\text{or}\qquad\frac{a}{b}\ge100 $$

When a problem is specified to have a solution it's usually meant that the problem has at least a solution, the exact number thereof can be greater than one.

The condition under which the given problem has one and only one solution is that the quadratic in $X$ has zero discriminant. Thus $b/a=1/100$ or $b/a=100$.