How to solve an exponential equation with two different bases: $3^x - 2^x = 5$

Can anyone tell me how to solve this equation $$3^x - 2^x = 5$$ other than graphically?

I'm stunned. I don't know what to do in the first step.

Solution 1:

Divide equation by $2^x$, then you get equation $$ \left(\frac{3}{2}\right)^x-1=\frac{5}{2^x} $$ Its left-hand side is strictly increasing, and right-hand side is strictly decreasing, hence there is at most one root. Obviously it is $x=2$.

Solution 2:

I assume you are interested only in real solutions. In general, $a^x-b^x=c$ for $a > b > 1$ and $c > 0$ will has a unique solution, as can be seen by looking at the function $f(x)=a^x-b^x-c$ and its derivative $f\,'(x)=a^x\ln a-b^x\ln b$. In particular, $f\,'(x)=0$ iff $x=x^*=-\frac{\ln\ln a-\ln\ln b}{\ln a-\ln b}<0$, and $f\,'$ is negative (and therefore $f$ is decreasing) for $x < x^*$ and positive (respectively, increasing) for $x > x^*$. So for $x < 0$, $f$ lies below $y=-c$, going from a horizontal asymptote (as $x\rightarrow-\infty$) to its global minimum at $x^*$, to $(0,-c)$ where it crosses the $y$-axis, while for $x > x^*$, $f$ is increasing and concave ($f\,''>0$). From its derivative $f\,'(0)=\ln a-\ln b$ at $0$, one can estimate the root of $f$ to be $x_1=\frac{c}{\ln a-\ln b}$, which always bounds (overestimates) the true root (since $f\,''>0$). Because of this concavity, the secant method with $x_0=0$ & $x_1$ above might give slightly better convergence than Newton's method for this root.

More generally, for $a,b,c>0$, the analysis & character of $f$ will vary depending on the signs of these quantities: $\chi_0=a-b$, $\chi_1=\ln a-\ln b$, $\chi_2=\ln\ln a-\ln\ln b$ which in turn depend on the signs of $\ln\frac{a}{b},~\ln a$ & $\ln b$ (i.e. on the order of $a$, $b$ and $1$ on the positive number line), thus leaving six cases to classify (sorry about the extra legend in the graph; not sure how it's getting there):

$$ \matrix{ \text{case}&\qquad\text{unique solution for}\\ a < b < 1&\qquad x < 0\\ a < 1 < b&\qquad x < 0\\ b < a < 1&\qquad\text{no solution}\\ b < 1 < a&\qquad x > 0\\ 1 < a < b&\qquad\text{no solution}\\ 1 < b < a&\qquad\text{discussed above, }x > 0\\ } $$

When $a$ is between $b$ and $1$, there is no solution. This can perhaps also be seen if one rewrites $f$ as $$ \eqalign{ f(x) &= a^x - b^x -c \\ &= 2 \, (ab)^{x/2} \sinh\left(\frac{x}{2}\ln\frac{a}{b}\right) - c \,. } $$

The cases when $c<0$ can then be inferred by interchanging $a$ and $b$, and of course $c=0$ has only the solution $x=0$ for $a\ne b$ both positive.

Solution 3:

I'm going to assume the real question is, given positive $a,b,c$, how to solve $a^x-b^x=c$. And I'm afraid the answer is, unless some simple answer jumps out at you (as $x=2$ for the original equation, $3^x-2^x=5$), the only way to solve it is numerically. Have you learned about Newton's Method yet?

Solution 4:

A modular view.

We can reformulate $$\small 3^x-2^x=3^2-2^2 \\ \small 3^x-3^2=2^x-2^2 \\ \small 3^2( 3^y-1)=2^2(2^y-1) $$ and finally a solution in integer terms $$\small { 3^y-1\over 2^2 }={ 2^y-1 \over 3^2 } $$ For the general case (real y) we see, that after y=0 and lhs=rhs=0 there is no other solution, since the lhs increases faster than the rhs if y increases above zero.

But we may look at it with modular arguments: then $\small 3^2 $ must be a factor of $\small 2^y-1 $ which we know is possible only if y is divisible by 6 (or zero), so we may replace $\small y=6z$. So we get $$\small { 9^{3z}-1 \over 2^2}={64^z-1\over 3^2} $$ But then the numerator of the lhs contains the primefactor 2 to the 3'rd power and its denominator only to the 2'nd power and the rhs is odd, so if not the lhs and rhs are simultaneously zero we have no solution in integers.

Thus it must be that $\small z=0 \to y=0 \to x=2 $ and no other solution

Solution 5:

There is also an insight via powerseries and series-inversion. The resulting power series seems to have convergence radius zero, but anyway. (It might be dealt by methods of divergent summation).
Define a function $$\small f(x) = \sum_{k=1}^{\infty} {\ln (3)^k - \ln(2)^k \over k!}x^k $$ This has no constant term and an exact (compositional) inverse power series can thus be defined: $$\small g(x)= f^{[-1]}(x) \\ \small \sim 2.46630346238 x - 5.44932534802 x^2 \\ \small + 17.9577561858 x^3 - 70.0338596112 x^4 \\ \small + 299.887672124 x^5 - 1362.93343700 x^6 + O(x^7) $$ Unfortunately, the (absolute value of the) quotient of subsequent coefficient seems to increase, so the convergence-radius is likely zero. However, using Borel- (or Noerlund) summation one can approximate values for $\small g(1) \sim _{(\mathcal B)} 1 $ , $\small g(5) \sim _{(\mathcal B)} 2 $ and $\small g(19) \sim _{(\mathcal B)} 3 $

(Note, that I didn't discuss multivaluedness here which should be considered in a full-featured answer)

[update]
Possibly the power series of g(x) has convergence for $\small \left| x\right|<1/6 $. We get using 128 terms of the power series $\small g(1/6) \sim 0.310913094947$ where then $\small w= g(1/6) $ and $\small 3^w-2^w \sim 1/6 $. Because the coefficients in $\small g(x) $ are alternating we can effectively sum using Cesaro/Euler-summation...