Solve the equation $$ \log_2(9x+2)=\log_3(16x+3). $$ It is easy to see that $x=0$ is a solution but how to prove that there are no more solutions?

My idea was prove that the function $f(x)=\log_2(9x+2)-\log_3(16x+3)$ is monotone for $x \geq 0$. We have $$ f'(x)=\frac9{ \left( 9\,x+2 \right) \ln \left( 2 \right) }- \frac{16}{ \left( 16\,x+3 \right) \ln \left( 3 \right) }, $$ and the inequality $f'(x)>0$ has the solution $$ \left( -2/9, -3/16 \right) \cup \left( a ,\infty \right), $$ where $$a=-{\frac {1}{144}}\,{\frac {32\,\ln \left( 2 \right) -27\,\ln \left( 3 \right) }{\ln \left( 2 \right) -\ln \left( 3 \right) }}$$ so $f'(x)>0$ for $x>0$. Thus $f(x)$ is there monotone and $f(x)=0$ has only one solution for $x \geq 0$.

My question Is there a solution without using the derivative?

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    $\begingroup$ What about $x = -\frac{1}{6}$? $\endgroup$ – Xander Henderson Apr 18 '18 at 18:57
  • $\begingroup$ @Xander Henderson, oh, I have lost one solution for $x<0$, thank you $\endgroup$ – Leox Apr 18 '18 at 19:05

If you are working with polynomials, it is relatively simple to determine whether or not you have found all of the solutions to an equation: once the number of solutions (counting multiplicity) is equal to the degree of the polynomial, you are done. Similar arguments can be generalized to equations involving rational functions. Unfortunately, once you get into transcendental functions (such as functions involving logarithms), you are almost certainly going to have to start using the tools of calculus.

For the problem you give, the basic approach is correct. There does seem to be one small error / bit of pedantry that should probably be mentioned: $\log(t)$ is typically only defined for $t > 0$, hence the first thing that I would do is ensure that I am working in the correct domain. We have $$ 9x+2 > 0 \iff x > -\frac{2}{9} \qquad\text{and}\qquad 16x + 3 > 0 \iff x > -\frac{3}{16}. $$ Combining these, we know a priori that $$x > -\frac{3}{16},$$ hence we don't even need to consider values of $x$ smaller than this.

The next bit of your argument looks fine: you note that if we define $$ f(x) := \log_2(9x+2) = \log_3(16x+3), $$ then $f'(x) > 0$ on the set $$ \left( -2/9, -3/16 \right) \cup \left( a ,\infty \right) $$ where $$a=-\frac1{144}\,\frac {27\,\ln \left( 3 \right)-32\,\ln \left( 2 \right) }{\ln \left( 3 \right) -\ln \left( 2 \right) }\approx-0.128$$ (I am assuming that your arithmetic is correct here—it seems reasonable). Note that the first interval is irrelevant because it is not in the domain of $f$, hence we can ignore it. It then follows from the intermediate value theorem that there can be at most one solution in the second interval (indeed, $x=0$ is that solution).

By a similar argument, $f$ is decreasing on the interval $$ \left( -3/16, a \right), $$ thus there can be at most one solution in that interval. Since $x=0$ solves the original equation and $f$ is increasing on an interval to the left of zero, we know that there is some $x' < 0$ such that $f(x') < 0$. We can also determine (using another tool that is generally introduced in calculus classes) that $$ \lim_{x\to-3/16} f(x) = +\infty. $$ By the intermediate value theorem, there must be another solution in this interval (indeed, $-\frac{1}{6}$ gets the job done). At this point, we have exhausted the domain of $f$, hence these are the only solutions.

Short version: When working with transcendental functions, it is unlikely that you will be able to say very much (in general) without resorting to results from calculus. These kinds of functions are, in some fundamental ways, really only understandable in the framework provided by limits, continuity, and the derivative.


HINT: Use the change of base formula to get$$\frac {\log(9x+2)}{\log 2}=\frac {\log(16x+3)}{\log3}$$Now isolate $x$


YES, we can solve this equation without using derivative. here I solved this equation by a method without using derivative.

$\log_2(9x+2)=\log_3(16x+3)=y $(say)

$\log_2(9x+2)=y \implies2^y=9x+2$

multiply this equation by 16$\implies2^{y+4}=144x+32$.........(1)

second part of given equation is $\log_3(16x+3)=y\implies 3^y=16x+3$

multiply this equation by 9 $\implies3^{y+2}=144x+27$.........(2)

$(1)-(2)\implies 2^{y+4}-3^{y+2}=5$.......(3)

obviously $y=1,-1$ is a solution

if $y=1\implies\log_2(9x+2)=\log_3(16x+3)=1$.By using the formula: if,$log_a(b)=1\implies b=a.$

similarly $y=-1$ we find $x=-1/6$

$9x+2=2\implies x=0$. similarly, $16x+3=3\implies x=0$

$x=0,-1/6$ is a solution . we want to prove $x=0,-1/6$ is only two solution

from (3), $2^{y+4}-3^{y+2}=5$,

if $y$ is greater than $1$, $2^{y+4}-3^{y+2}$,is negative.i.e,$2^{y+4}\lt3^{y+2}$.we prove it by induction

it is true for $y=1$,let it is true for $y=k$.

i.e,$2^{k+4}\lt3^{k+2}$ we know that $2\lt3$

multiply these two inequalities,we get $2^{(k+1)+4}\lt3^{(k+1)+2}$, by $PMI$ it is true for all natural numbers greater than $1$.

for $y\gt1$,there is no solution for$2^{y+4}-3^{y+2}=5$,since , $2^{y+4}-3^{y+2}$ is negative for $y\gt1$. therefore, there is no corresponding value for $x$.

Therefore $\log_2(9x+2)-\log_3(16x+3)$ has no solution other than $x=0,-1/6$

  • $\begingroup$ my answer is correct. please read fully. if ,x has another solution in rational number . I requested you to ''please show it''. $\endgroup$ – NEW SUN May 12 '18 at 14:26
  • $\begingroup$ This shows that $y = 1$ is the only positive integer that solves $2^{y+4}-3^{y+2}=5$. What about all other possible values of $y$? Nothing in the problem says $x$ or $y$ has to be an integer. It turns out that if $y>1,$ then $2^{y+4}-3^{y+2}<5$ even if $y$ is not an integer, so there are no solutions for $y>1$ and therefore no solutions for $x>0.$ But $2^{-1+4}-3^{-1+2}=5$, and therefore there is a second solution, $y=-1$ and $x = -\frac16$. $\endgroup$ – David K May 12 '18 at 14:29
  • $\begingroup$ Note that $x=-\frac16$ was already given in another answer when I started commenting, but you have to read that answer to the end to find it. $\endgroup$ – David K May 12 '18 at 14:31
  • $\begingroup$ By the way, nowhere in the problem does it say $x$ has to be rational. It happens that the second solution is rational, but if it had been irrational it still would be a second solution. $\endgroup$ – David K May 12 '18 at 14:33
  • $\begingroup$ $x = -\frac16$ is also given in the first comment under the question. $\endgroup$ – David K May 12 '18 at 14:36

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