How to solve this logarithmic equation?

Question

The equation $$x^2\cdot\log_2\dfrac{3+x}{10}-x^2\cdot\log_{\frac{1}{2}}(2 + 3x)=x^2 - 4+2\cdot\log_{\sqrt{2}}\dfrac{3x^2+11x+6}{10}$$ has two solutions for x, namely 1 and 2. I tried

Reduce[x^2*Log[2, (3 + x)/10] - x^2*Log[1/2, 2 + 3*x] == 
  x^2 - 4 + 2*Log[Sqrt[2], (3*x^2 + 11*x + 6)/10], x]

and

Solve[x^2*Log[2, (3 + x)/10] - x^2*Log[1/2, 2 + 3*x] == 
  x^2 - 4 + 2*Log[Sqrt[2], (3*x^2 + 11*x + 6)/10], x]

but I could not get these two solutions. How do I tell Mathematica to do that?

Answer 1

Here's a way to do it numerically. It uses the adaptive sampling Plot uses to determine the approximate positions of the roots.

Here's the expression:

expr = x^2*Log[2, (3 + x)/10] - 
   x^2*Log[1/2, 2 + 3*x] - (x^2 - 4 + 
     2*Log[Sqrt[2], (3*x^2 + 11*x + 6)/10]);

We'll solve for expr==0. We look for roots between

xlow = 0; xhigh = 3;

To do that, ask Plot to plot the expression, then extract the list of data points:

(points = 
   Cases[Normal[Plot[expr, {x, xlow, xhigh}]], Line[l_] :> l, 
     Infinity] // First) // Shallow

These are the points Plot uses. You can ListPlot them to see the adaptive sampling at play.

Now, we need to locate which two points bracket a root, which means that their y-coordinates have opposite signs; so check for this:

signChanges = Position[
   Times @@ ## & /@ Partition[points[[All, 2]], 2, 1],
   x_ /; x < 0
   ] // Flatten

{23, 232}

and finally use FindRoot giving it a range to search in:

FindRoot[expr == 0, {x, points[[#, 1]], points[[# + 1, 1]]}] & /@ signChanges
{{x -> 1.}, {x -> 2.}}

EDIT: changed the way the points are obtained after discussion with Artes and JM

Answer 2

This equation has many more 'nice' valued roots than you think. Let's see what a man-machine cooperation can find. I'm going to take fairly rough steps and back-substitute the results to see whether I stepped on any mathematician's toes in the process.

First, let's move everything to one side of the equation and whisk everything on that side together:

 (Exp[x^2*Log[2, (3 + x)/10] - x^2*Log[1/2, 2 + 3*x]] - 
  Exp[x^2 - 4 + 2*Log[Sqrt[2], (3*x^2 + 11*x + 6)/10]]) // Together

(1)

Since it is easy to see that (3+x)(2+3*x) == (6 + 11 x + 3 x^2) the right hand part of the above equation equals zero if (6 + 11 x + 3 x^2)==0.

Solve[6 + 11 x + 3 x^2 == 0, x]

{{x -> -3}, {x -> -(2/3)}}

Let's further explore that part of the numerator of (1). We take the log of both parts (allowed because the whole term between parenthesis should equal zero, add a few minus signs to make everything still zero). Rewrite the logs using a few rules for logarithms:

  Log[-10^((4/Log[2])) E^4 (3 + x)^(x^2/Log[2]) (2 + 3 x)^(x^2/Log[2])] - 
  Log[-10^((x^2/Log[2])) E^x^2 (6 + 11 x + 3 x^2)^(4/Log[2])] 
   //. {Log[a_ b_] -> Log[a] + Log[b], Log[a_^b_] -> b Log[a]}

Again, looking through your eye lashes you notice you can factor out a term (4-x^2):

The first part has solutions $\pm 2$, of course, and for the second part we find:

Solve[ 1 + Log[10]/Log[2] - Log[6 + 11 x + 3 x^2]/Log[2] == 0, x]

{{x -> -(14/3)}, {x -> 1}}

Back-substituting all solutions in your equation:

x^2*Log[2, (3 + x)/10] - x^2*Log[1/2, 2 + 3*x] == 
x^2 - 4 + 2*Log[Sqrt[2], (3*x^2 + 11*x + 6)/10] 
  /. {{x -> -14/3}, {x -> -2}, {x -> 2}, 
      {x ->  1}, {x -> -3}, {x -> -2/3}} // FullSimplify

{False, True, True, True, True, True}

So, we have solutions for x = -3,-2, -2/3, 1, and 2.

However, as acl, Artes and Whuber have pointed out, some of these have a problem. For x =-3 we get a Log[2,0] in the first term of the left-hand side of the equation that Mathematica cancels with another Log[0]; same for the second term and x = -2/3. The situation for x=-2 is different. You get logs of negative numbers like Log[-4] which have a complex answer that depending on the branch cut definition that is being used. In this case we get I Pi + Log[4]. If you do Exp[I Pi + Log[4]], you get -4. So, depending on your stance on branch cuts you might consider the full set of solution to be {-2, 1, 2}. And if you don't like the first one, we've still found 1 and 2 as exact values, not with numerical root finding approximations.

Answer 3

The most reliable approach to the problem uses Reduce, one can also use Solve. However one should restrict the complex domain to a compact subset or should add the domain specifications like e.g. Reals, remembering that Mathematica treats numbers as complex in general.

In Mathematica 7 Root had been considerably updated to include these capabilities:

Root[{f, x0}] represents an exact root of the general equation f[x]==0, which can be transcendental.

In Root[{f, x0}], x0 must be an approximate real or complex number such that exactly one root of f[x] lies within the numerical region defined by its precision.

Root[{f, x0, n}] represents n roots, counting multiplicity, that lie within the numerical region defined by the precision of x0.

Let's define f :

f[x_] := x^2 Log[2, (3 + x)/10] - x^2 Log[1/2, 2 + 3 x] - x^2 + 4 
        - 2 Log[Sqrt[2], (3 x^2 + 11 x + 6)/10]

Reduce[ f[x] == 0, x, Reals]

x == Root[{-4Log[2] + Log[(6 + 11#1 + 3 #1^2)^4/10000] + Log[20/((3 +#1) (2 + 3#1))] #1^2 &,
             0.999999999999999999998}] || 
x == Root[{-4Log[2] + Log[(6 + 11#1 + 3#1^2)^4/10000] + Log[20/((3 + #1) (2 + 3#1))] #1^2 &,
             2.0000000000000000000}]

N[ %, 30]

x == 1.00000000000000000000000000000 || x == 2.00000000000000000000000000000

We can find solutions in appropriately restricted complex plain, but then neither Reduce nor Solve can prove that the solution set is complete :

Solve[ f[x] == 0 && -5 < Re[x] < 5 && -5 < Im[x] < 5, x]

Solve::incs: Warning: Solve was unable to prove that the solution set found is complete. >>

{{x -> 2},
 {x -> Root[{-4Log[2] + 4Log[1/10 (6 + 11 #1 + 3 #1^2)] + Log[2] #1^2 
             - Log[(3 + #1)/10] #1^2 - Log[2 + 3 #1] #1^2 &,
  1.000000000000000000000000000000}]}}

One might wonder that using e.g. f[-2] // Simplify yields 0 but there is no root -2 in the output of Reduce, even though we can see this :

Plot[{ Re @ f[x], Im @ f[x]}, {x, -5, 5}, PlotStyle -> Thick, Evaluated -> True]

The reason is that logarithm is not well defined in the whole complex plain, there must be certain branch cut or more generally one should view it as a function on a Riemann surface. Therefore we should rely on the Reduce output rather than on Simplify.

Answer 4

Taking to heart the advice given by J.M. in a comment to the question, after finding that the brute-force method of just asking for a solution falls short, we might consider using Mathematica as a tool to support the operations we would think of carrying out when solving such an equation.

Staring at the equation suggests several methods might be fruitful:

1. Subtracting one side from the other, thereby enabling one expression to be simplified fully.

2. Perhaps finding common factors among the polynomial subexpressions. This would primarily involve factoring the highest-degree polynomial, $3x^2 + 11x + 6$.

3. Exploiting the multiplicative property of logarithms, $\log(uv)=\log(u)+\log(v)$.

$Mathematica$ could use a little nudging in all these directions.

Developing solutions

Begin with the original equation:

f[x_] := x^2*Log[2, (3 + x)/10] - x^2*Log[1/2, 2 + 3*x] == 
         x^2 - 4 + 2*Log[Sqrt[2], (3*x^2 + 11*x + 6)/10]

$\frac{x^2 \log \left(\frac{x+3}{10}\right)}{\log (2)}+\frac{x^2 \log (3 x+2)}{\log (2)}=x^2+\frac{4 \log \left(\frac{1}{10} \left(3 x^2+11 x+6\right)\right)}{\log (2)}-4$

Apply the three rules. We will be looking for zeros of g:

g[x_] := Evaluate[Replace[Together[f[x] /. u_ == v_ -> u - v] , u_Plus :>  Factor[u], Infinity] 
         //. Log[u_ v_] -> (Log[u] + Log[v]) // Simplify]

$\frac{\left(x^2-4\right) \left(\log \left(\frac{x+3}{20}\right)+\log (3 x+2)\right)}{\log (2)}$

It looks a lot simpler (you can practically read all the solutions directly from this form), so let's try with Solve once more:

Solve[g[x] == 0, x, Reals]

{{x -> 1}, {x -> 2}}

If solutions found using Complex arithmetic are desired, unfetter Solve by removing the last option:

Solve[g[x] == 0, x]

{{x -> -2}, {x -> 1}, {x -> 2}}

As a check, one can plot $g$ in the neighborhood of the putative solutions and look for crossings of the x-axis.

Plot[g[x], {x, -3, 3}]

---

Checking the solutions

Surprisingly, a direct check of the solutions fails:

f /@ {-2, 1, 2}

{False, False, False}

Tracing the function identifies the problem:

Trace[f[#]][[-2 ;;]] & /@ {-2, 1, 2} 

$\left\{\left\{\frac{4 (i \pi +\text{Log}[4])}{\text{Log}[2]}-\frac{4 \text{Log}[10]}{\text{Log}[2]}==\frac{4 \left(i \pi -\text{Log}\left[\frac{5}{2}\right]\right)}{\text{Log}[2]},\text{False}\right\},\left\{-\frac{\text{Log}\left[\frac{5}{2}\right]}{\text{Log}[2]}+\frac{\text{Log}[5]}{\text{Log}[2]}==1,\text{False}\right\},\left\{8==\frac{4 \text{Log}[4]}{\text{Log}[2]},\text{False}\right\}\right\}$

For some reason, the penultimate expressions do not evaluate correctly. Although other threads have been related to similar problems with logarithms (such as simplifying $\frac{\log x^a}{a} = \log x$), they do not appear fully to discuss why expressions like, say, 3 == Log[8]/Log[2] evaluate to False whereas (3 - Log[8]/Log[2] // Simplify) == 0 will evaluate to True. This apparent contradiction casts doubt on the easy--but probably not fully correct--explanation that the logarithm is a many-valued function.