How to solve the equation:

$$2\cdot 3^x +2^{2x}+5^{2x-1}-13^x+10=0$$

Well the answer can be found by trial & error to be $x=2$. But I am not able to proceed in a systematic way.

I cannot see taking a factor out to solve. Is there any general algorithms for these kind of problems?

  • 1
    $\begingroup$ In general, there isn't a method for equations like these. Sometimes clever solutions can be pulled out, but with all these primes, I highly doubt there's a solution to this. $\endgroup$ – Kaynex Mar 13 '16 at 18:09
  • $\begingroup$ There are two solutions quite close to each other on the real line, $2$ exactly and a transcendental number near $1.88537$. $\endgroup$ – dbanet Mar 13 '16 at 18:13
  • $\begingroup$ @dbanet How do you know it is transcendental? $\endgroup$ – Bobson Dugnutt Mar 13 '16 at 20:15

Let's start by seeing if we can find bounds on where a solution can be. We start by naming the object we want to study. $$ f(x) = 2 \cdot 3^x + 2^{2x} + 5^{2x-1} - 13^x + 10 $$ You would like to know when $f(x) = 0$. Exponentials behave differently if their exponents are positive or negative, so it is useful to find out where those are. If we simultaneously require $x > 0$, $2x > 0$, and $2x -1 > 0$, we find $x > 1/2$. In the other direction, $x < 0$, $2x < 0$, and $2x -1 < 0$, we find $x < 0$. So, in some sense the analysis will be easier for $x < 0$ and $x > 1/2$ than for $x \in [0,1/2]$. Let's first check $x = 0$ and $x = 1/2$ so we only need to consider open intervals. \begin{align*} f(0) = 2 \cdot 1 + 1 + 1/5 - 1 + 10 &= 61/5 \neq 0 \\ f(1/2) > 2 \cdot 1 + 2 + 1 - 4 + 10 &= 10 > 0 \text{,} \end{align*} where for the second line we have used that $3^{1/2} > 3^0 = 1$ and $9 < 13 < 16$ so that $3 < 13^{1/2} < 4$, so even if we add too little and subtract too much, we still aren't down to zero. (If we think about what we've done -- add up the smallest versions of the terms as $x$ ranges over $0$ to $1/2$ and subtract the largest (in magnitude) negative term over that range and still the sum is positive -- we've actually shown that $f$ is never zero in $[0,1/2]$. In fact, we can keep going to $x=1$: $$ f(1) > 2 \cdot 1 + 1 + 1 - 13 + 10 = 1 > 0 \text{.} $$)

Now let's use a similar "sloppy" (i.e. approximate things with easy approximations) argument to show that we need not consider large $x$s. Suppose $x \geq 3$. All the exponents are positive. We expect the sign of $f$ to depend on what $5^{2x-1}$ and $13^x$ are doing (because

  • $2 3^3$ = 54 and only grows by a factor of $3$ every time $x$ increases by $1$,
  • $2^{2 \cdot 3} = 64$ and only grows by a factor of $4$ every time $x$ increases by $1$,
  • $5^{2\cdot 3 - 1} = 3125$ and increases by a factor of $25$ every time $x$ increases by $1$, and
  • $13^3 = 2197$ and only increases by a factor of $13$ every time $x$ increases by $1$.

So the negative summand is smaller than the power of $5$ when $x = 3$ and the gap only increases as $x$ increases. We could estimate the first two summands with $0$ and the rest of the terms would still show that $f(x) > 0$ when $x \geq 3$. So if there is a solution with positive $x$, it occurs when $1 < x < 3$. In fact, we can use these observations to push the upper bound down a little : When does the "dominating" term equal the negative term? That is find $x$ in $(1,3)$ where $5^{2x-1} = 13^{(2x-1) \log_{13}(5)} = 13^x$. This is where $(2x-1) \log_{13}(5) = x$. This is easy algebra and we find that this happens when $x = \dfrac{\log_{13}(5)}{-1 + 2 \log_{13}(5)}$. We can estimate this as we have done above and find that this value is a little less than $5/2$. So any zero of $f$ has $x$ in $(1,5/2)$.

Suppose $x$ is negative and huge. We expect the powers to be almost zero and only the $10$ is left. When $x = 0$, the powers are as big as they're ever going to be (i.e. either $1$ or $1/5$), we can lower bound the positive summands with $0$ and upper bound the negative summand by $1$, and we find $f(x) \geq 0+0+0-1+10 = 9$ for any $x < 0$. That is, there is no solution with $x < 0$. Therefore, all solutions to the equation have $1 < x < 5/2$.

This is usually how one starts attacking a problem in many exponentials -- use very easy estimates on lower bounds and upper bounds to eliminate large intervals where a solution cannot be (because one summand is "winning" so the sum cannot be near zero). Having done this, the rest of the problem tends to be a little more difficult.

On the interval $(1,3)$, all the exponents are positive and we are watching a "fight" between the positive summands, keeping the sum positive, and the negative summand, trying to overcome their resistance and drag the sum negative. (If it ever succeeds, since $f$ is continuous, we will get at least one zero of $f$. We might get two or more, depending how far past zero that negative term can drag the sum and whether the terms we have been disregarding because they are small outside the interval $(1,5/2)$ contribute to "wiggling" in the interval.)

There's only one integer in the interval, so before we try difficult things, we should try that integer. $$ f(2) = 2 \cdot 9 + 16 + 125 - 169 + 10 = 0 \text{.} $$ So, (shock) there does seem to be a zero of $f(x)$ at $x = 2$. So we should factor that one out before attempting anything complicated, like a numerical method. We now consider finding the roots of $$ g(x) = \frac{f(x)}{x-2} \text{.} $$ Checking at $1$ and $5/2$, we find (using estimates as above for the second line) \begin{align*} g(1) &= -12 \text{ and } \\ g(5/2) &> 0 \text{,} \end{align*} so there is another zero in the interval $(1,5/2)$. By plugging in values near the midpoint of the surviving interval (i.e., by using the bisection method), we find another root of $g(x)$ and hence of $f(x)$ at $x$ in the interval $(9/5, 2)$. (We can approximately factor out this root via $g(x)/(x-1.9)$ and the result is greater than $10$ on the interval $(1,5/2)$ except possibly for another root between $1.87$ and $1.9$ and using methods from calculus, we can show that this root is an artifact of the approximation in dividing by $x-1.9$ instead of by the correct value: $x-1.88537\dots$.

Thus we have found both of the roots of $f$, one by a lucky guess (after eliminating all the other easy choices) and one by a numerical method.


I believe you can only solve problems like this numerically. For example, you could use the Newton-Raphson Method which states that if you have a function defined as:$$f(x)=0$$then the solutions can be iterated towards using the folowing:$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$where $f'(x_n)$ represents the differential of $f(x)$ with respect to $x$ evaluated at $x=x_n$.

You would start by guessing some initial value for $x=x_0$ and the iteration above would iterate towards one of the roots of that equation.

Plugging your equation in wolframalpha gives two solutions $x=2$ and $x\approx1.88537$


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