Find all real solutions of $\log_2 x + \log_x 9 = \log_2 (x+9)$ without using derivatives

Question

Find all real solutions of the logarithmic equation $$\log_2 x + \log_x 9 = \log_2 (x+9)$$ without using derivatives.

My teacher said that this problem can be solved using only basic facts on logarithmic functions and increasing/decreasing functions, so there is no need to use derivatives to verify all solutions.

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For someone who is interested in a solution using derivatives, we note that $x > 0$ and $x \ne 1$. By direct manipulations and substituting $x = 3^t$, we get $$ \begin{align} \log_2(3^t) + \log_{3^t}(9) &= \log_2(3^t+9) \\ \frac{2}{t} &= \;\log_2 \left(\frac{3^t+9}{3^t} \right) \\ t &= \frac{2\ln2}{\ln(1+3^{2-t})}. \end{align} $$ Letting $f(t) = \frac{2\ln 2}{\ln(1+3^{2-t})}$, we find that $$ f''(t)=\frac{2(\ln 2)(\ln 3)^2(18-3^t\ln(1+3^{2-t}))}{(3^t+9)^2(\ln(1+3^{2-t}))^3}. $$ We use the fact that $\ln(1+x) \le x$ for all $x \in \mathbb{R} \implies 18-3^t\ln(1+3^{2-t}) \ge 18-3^t(3^{2-t}) = 9>0$. Therefore, $f''(t) > 0$ for all $t \in \mathbb{R}$, and we can conclude that $f$ is strictly convex on $\mathbb{R}$.

Since $f$ is strictly convex, the equation $f(t)=t$ must have at most $2$ solutions. By inspection, $t \in \{1,2\}$ satisfies this equation. Hence, the equation $\log_2 x + \log_x 9 = \log_2 (x+9)$ has $2$ solutions, which are $x \in \{3^1,3^2\} = \{3,9\}$.

Answer 1

The following is simply a speculation that this is not possible

It appears that any such proof may simply be a reformulation of various ideas related to derivatives (and $n$th derivatives in general).

Consider that we must, at some stage, have the following cases:

$1 \leq x < 3, 3 <x< 9,$ and $x >9$.

These intervals are required to analyze the behavior of the functions near the exact solutions.

$\log_2 x + \log_x 9 = \log_2 (x+9) \implies \log_2\frac{x}{x+9} +\log_x9 = 0.$

Heuristically, one may propose that the cases we considered imply the following, respectively: $\log_2\frac{x}{x+9} +\log_x9 > 0, \log_2\frac{x}{x+9} +\log_x9 < 0,\log_2\frac{x}{x+9} +\log_x9 > 0$.

Thus, as an example, we may consider $x > 9$. We want to somehow show that $x>9 \implies \log_2\frac{x}{x+9} +\log_x9 > 0$.

Notice that $\log_2\frac{x}{x+9}< 0, \log_x9 >0$, and that both functions approach $0$.

In essence, the only way to show the sum is nonzero on this interval, and on any entire interval, is to show that one of them approaches $0$ faster than the other, which is equivalent to using derivatives.

Answer 2

This is an unfinished draft. Perhaps it contains some ideas that someone else can use towards a complete proof. So far this draft shows that any solution must satisfy $e<x\leq9$.

Let $x$ be a positive real number such that $$\log_2(x)+\log_x(9)=\log_2(x+9).\tag{1}$$ For $\log_x(9)$ to make sense we must have $x\neq1$. We can convert to natural logarithms by noting that $$\frac{\ln(9)}{\ln(x)}=\log_x(9)=\log_2(x+9)-\log_2(x)=\frac{\ln(x+9)-\ln(x)}{\ln(2)},$$ and hence we have $$\ln(x)\Big(\ln(x+9)-\ln(x)\Big)=\ln(2)\ln(9).\tag{2}$$ To find all solution without using derivatives, it matters to know your definition of the natural logarithm for positive reals. I'll take it to be the unique inverse of the exponential function $x\ \mapsto\ e^x$. I'll use the following two facts about the exponential function, proofs can be found elsewhere on this forum:

Observation 1: The function $f(t):=\left(1+\frac1t\right)^t$ is strictly increasing with $\lim_{t\to\infty}f(t)=e$.

Proof. See for example here.$\qquad\square$

Observation 2: The function $g(t):=\left(1+\frac1t\right)^{t+1}$ is strictly decreasing with $\lim_{g\to\infty}f(t)=e$.

Proof. See for example here.$\qquad\square$

Lemma 3: For all $t>0$ we have $$\ln(t+1)-\ln(t)<\frac1t.$$

Proof. By elementary algebraic manipulation this is equivalent to $$\ln\left(1+\frac1t\right)<\frac1t \qquad\text{ and hence to }\qquad e>\left(1+\frac1t\right)^t, $$ which holds for all $t>0$ by Observation 1. $\qquad\square$

Substituting $t=\tfrac x9$ into equation $(2)$ we get $$\ln(2)\ln(9)=\ln(9t)\Big(\ln(t+1)-\ln(t)\Big),$$ and hence applying Lemma 3 yields $$\ln(2)\ln(9)<\frac{\ln(9t)}{t}, \qquad\text{ or equivalently}\qquad \frac{\ln(x)}{x}>\frac{\ln(2)\ln(9)}{9}.$$

Lemma 4: The function $\frac{\ln(x)}{x}$ is strictly increasing for $x<e$, and strictly decreasing for $x>e$.

Proof. Let $x>y>e$. Then we want to show that $\frac{\ln(x)}{x}<\frac{\ln(y)}{y}$, or equivalently $x^y<y^x$. Writing $x:=(1+c)y$ with $c>0$, it suffices to shows that $$\left((1+c)y\right)^y<y^{(1+c)y},\qquad\text{ or equivalently }\qquad y>(1+c)^{\tfrac1c}.$$ With Observation 1 this follows immediately from the fact that $y>e$. The proof for $y<x<e$ is entirely analogous.$\qquad\square$

This shows that the left hand side of equation $(2)$ is monotonically increasing for $x<e$, because $$2^{\ln(9)}=\left(1+\frac9x\right)^{\ln(x)}=\left(\left(1+\frac9x\right)^{\frac{x}{9}}\right)^{9\frac{\ln(x)}{x}}.$$ The base $\left(1+\frac9x\right)^{\frac{x}{9}}$ is strictly increasing by Observation 1, and the exponent is strictly increasing for $x<e$ by Lemma 4. So for $x\leq e$ we have $$\left(1+\frac9x\right)^{\ln(x)}\leq \left(1+\frac9e\right)^{\ln(e)}=1+\frac9e.$$ It is a matter of elementary algebra to show that $2^{\ln(9)}>1+\tfrac9e$, and so we conclude that any solution to equation $(1)$ must have $x>e$.

Similarly we can obtain a better upper bound by using Observation 2 and a slight variation on Lemma 4:

Lemma 5: The function $\frac{\ln(x)}{x+9}$ is strictly decreasing for $x>9$.

Proof. Let $x>y>9$. Then we want to show that $\frac{\ln(x)}{x+9}<\frac{\ln(y)}{y+9}$, or equivalently $x^{y+9}<y^{x+9}$. Writing $x:=(1+c)y$ with $c>0$, it suffices to shows that $$\left((1+c)y\right)^{y+9}<y^{(1+c)y+9},\qquad\text{ or equivalently }\qquad y^{\tfrac{y}{y+9}}>(1+c)^{\tfrac1c}.$$ By Observation 1 we have $(1+c)^{\tfrac1c}<e$ and it is clear that $y^{\tfrac{y}{y+9}}$ is strictly increasing for $y>1$. For $y=9$ we have $y^{\tfrac{y}{y+9}}=3>e$, which completes the proof.$\qquad\square$

Now as before we can decompose the left hand side of $(2)$ as follows: $$2^{\ln(9)}=\left(1+\frac9x\right)^{\ln(x)}=\left(\left(1+\frac9x\right)^{\frac{x+9}{9}}\right)^{9\frac{\ln(x)}{x+9}}.$$ As before, the base $\left(1+\frac9x\right)^{\frac{x+9}{9}}$ is strictly decreasing by Observation 2, and the exponent is strictly decreasing for $x>9$ by Lemma 5. For $x=9$ we of course find that $\left(1+\frac9x\right)^{\ln(x)}=2^{\ln(9)}$, so this shows that there are no solutions with $x>9$.

Some remarks

1. The proof of Lemma 5 shows that we can improve this upper bound for $x$ a bit; we readily see that $y^{\tfrac{y}{y+9}}>e$ for $y=9$, but it is also clear that this is true for some smaller values of $y$.
   With the help of a computer this goes down to $y\approx8.2$ and so
   the only solution with $x>8.2$ is $x=9$.
2. The proof of Observation 2 can be sharpened to show that $h(t)=(1+\tfrac1t)^{t+\tfrac12}$ is strictly decreasing, again with limit $e$ of course. Then we can use the same argument as in the proofs of Lemma's 4 and 5 for the function $\frac{\ln(x)}{x+9/2}$, to show that it is strictly decreasing for $x>6$ (or even $x>5.86$ with the help of a computer), and hence conclude that $x=9$ is the unique solution with $x>6$.

Answer 3

$\newcommand{\deq}{\stackrel{\triangle}{=}}$ $\DeclareMathOperator{\sgn}{sgn}$ Write $$g(t)\deq(1+3^{2-t})^t\text{.}$$ Evidently, $g(t)<1$ if $t<0$ and $g(t)=(1+3^{2-t})^t <1 + \frac{t3^{2-t}}{1-(t-1)3^{2-t}}=1 + \frac{t}{3^{t-2}-(t-1)} <1 + \frac{t}{1+2t-4-t+1}= 1 +\frac{t}{t-2}<4$ if $t>3$ by Bernoulli's inequality, so that the goal is to show that $$\begin{align} g(t)&>4 & t&\in(1,2)\text{,} \\ g(t)&<4 & t&\in(0,1)\cup(2,3)\text{.} \end{align}$$ Our strategy is to produce a rational approximation $h(t)$ to $g(t)$ such that

$$\begin{align} g(t)&>h(t) & t&\in(1,2)\text{,} \\ g(t)&<h(t) & t&\in(0,1)\cup(2,3)\text{,} \end{align}$$ then to use a computer algebra implementation of Vincent's theorem to decide whether $$\begin{align} h(t)&>4 & t&\in(1,2)\text{,} \\ h(t)&<4 & t&\in(0,1)\cup(2,3)\text{.} \end{align}$$

To produce such an approximation without differential calculus, we invoke the following fact:

The function $\digamma^x(s)\deq (1+x)^s$ is strictly absolutely monotone if $x>0$ and strictly completely monotone if $x<0$.

(A function $s\mapsto \phi(s)$ is said to be strictly absolutely monotone if all of its divided differences $\Delta^n\phi[s_0,s_1,\ldots,s_n]$ at pairwise-distinct inputs $\{s_0,s_1,\ldots,s_n\}$ are strictly positive. It is strictly completely monotone if $s\mapsto \phi(-s)$ is strictly absolutely monotone. Strict absolute monotonicity implies being positive, strictly increasing, and strictly convex.)

The consequence of this fact that we use is that the convergent $p^n_x(s)$ of degree $n-1$ of the Newton series of $\digamma^x(s)$ $$p^x_n(s)\deq \sum_{k=0}^{n-1}x^k\binom{s}{k} $$ has a remainder of predictable sign: $$\begin{align}\frac{(\sgn x)^n}{\binom{s}{n}}\left(\digamma^x(s)-p^x_n(s)\right)\propto (\sgn x)^n\Delta^n\digamma^x[s,0,\ldots,n-1] > 0\text{.} \end{align}$$

So define $$h(t)\deq \left. \frac{p_6^z(t+2)}{(1+z)^2}\right\rvert_{z=\tfrac{1}{3}p_4^2(3-t)}\text{.}$$

It follows, then, that $h$ bounds $g$ in the "right way" as described above, as this plot illustrates:

$g(t)$ and $h(t)$" />

Once the CAS has checked that $h(t)>4$ on $t\in(1,2)$ and $h(t)<4$ on $t\in(0,1)\cup(2,3)$ the demonstration is finished.