Exponential Equation with a Unique Solution

Suppose $y = 3^x > 0$ . Then our equation will be: $ry + 1/y = 3 \Rightarrow ry^2 - 3y + 1 = 0 \Rightarrow y = \frac{3 \pm \sqrt{9-4r}}{2}$ . Now, to avoid imaginary numbers, $9-4r \geq 0$ . If the discriminant is greater than zero, then the equation will have two positive solutions since ( $sqrt{9-4r}$ is less than 3 for positive $r$ , so both values of $y$ will be positive, so there will be two solutions for $x$ ). Hence, to be unique, the discriminant must be 0. So $r = \frac{9}{4}$ .

[Edits for clarity - Calvin]

Solution 1: Consider the substitution $y = 3^x$ , where $y = 3^x > 0$ . The given equation becomes $ry^2 -3y + 1 =0$ , which has roots $\frac {3 \pm \sqrt{9-4r} } {2r}$ . This would lead to 2 real solutions of $x$ unless $9-4r = 0$ or $\frac {3 - \sqrt{9-4r} } {2r} < 0$ . In the first case, we have $r= \frac {9}{4}$ , and in the second case, since $r>0$ , $3 - \sqrt{9-4r} > 3-3 =0$ so the term is always positive. Thus, there is a unique solution only when $r = \frac {9}{4}$ . Hence $a+b = 9+4 = 13$ .

Solution 2: Consider the graph of $f(x) = 3^{1-x} - 3^{-2x}$ . Then, $r 3^x + 3^{-x} =3 \Leftrightarrow f(x)=r$ , so we seek values $r$ where $f(x)=r$ has exactly 1 solution.

Let $z=3^{-x}$ , then $f(x) = -z^2 +3z = - (z - \frac {3}{2})^2+\frac {9}{4}$ . Let $x^* = -\frac{ \ln ( \frac {3}{2} )} {\ln (3)} = \frac{ \ln ( \frac {2}{3})}{\ln(3)}$ . The transformation shows that the graph of $f(x)$ in the domain $(-\infty, x^*]$ is increasing from $-\infty$ to $\frac {9}{4}$ , and the graph of $f(x)$ in the domain $[x^*, \infty)$ is decreasing from $\frac {9}{4}$ to $0$ . Thus, $f(x)=r$ has exactly 1 solution only when $r = \frac {9}{4}$ . Hence $a+b=9+4=13$ .

Let 3^x = y

implies we've yr + 1/y = 3

ry^2 + 1 = 3y

ry^2 - 3y + 1 = 0

y=(3 +/- sqrt(9-4r))/2r

since the original equation has a unique solution

9-4r = 0

hence r =9/4

For r = a/b we are asked to find a+b

hence a+b=9+4 = 13

Because f(x)=r*3^x + 3^-x looks similar to a parabola, if it is to equal 3 only once, it must be at its minimum!

first, we derive f to get f'(x)=r ln(3) 3^x-ln(3) 3^-x. Then, we set f'(x)=0 to get r ln(3) 3^x-ln(3) 3^-x=0 => r ln(3) 3^x=ln(3) 3^-x => r 3^x=3^-x p.t. r≠0 => 3^(2x)=1/r => 2x=log3(1/r) => x=log3(1/r)/2=log3(1/r^(1/2))

now that we have it's minimum in terms of r, if we plug it into f(x) and set f(x)=3, we'll have found out r. r 3^(log3(1/r^(1/2)) + 3^-(log3(1/r^(1/2))=r 1/r^(1/2) + r^(1/2)=2r^(1/2)=3 => 4r=9 => r=9/4.

9 and 4 are coprime and 9+4=13.

$\text{Since } 3^x \neq 0, \text{ we multiply the given equation by } 3^x \text{ to obtain} \\ \text{the 2nd degree equation } \\ \text{ } \\r(3^{2x}) - 3\cdot3^x +1 = 0 \\ \text{ } \\ \text{This equation will have a unique solution if the} \\ \text{descriminant = 0, i.e.} \\ \text{ } \\9-4r=0 \\ \text{ } \\ \text{so r = 9/4.} \\ \text{ } \\ \text{[Note that since } 3^x \text{ is a purely increasing function, a unique} \\ \text{solution for } 3^x \text{leads to a unique solution for x.]} \\ \text{ } \\ \text{We need to make sure that this leads to a valid solution for x,} \\ \text{ i.e that } 3^x>0. \text{ If } r=9/4, \text{we get the equation} \\ \text{ } \\ 9/4(3^{2x}) -3\cdot3^x+1=0 \\ \text{ } \\ \text{which gives us} \\ 3^x=2/3.$

Multiply 3^x on both sides of the equation to obtain r3^(2x) + 1 - 3^(x+1)=0. Let y = 3^x. A quadratic equation is obtained. ry^2 -3y + 1 = 0 Since the solution to the above equation is unique, the discriminant=(-3)^2 -4r= 0. r= 9/4. a+b=13.

Express the equation r(3^ x) + 3^ (-x) = 3 as quadratic equation by simplifying negative exponent, hence,

r(3^x) + 3^(-x) = 3

\Rightarrow r(3^ {2x}) - 3 (3^x) +1 = 0.

Since the equation has a unique real solution, then we can use the discriminant of the quadratic formula, i.e.,

D = b^2 - 4ac = 0, where a= r, b= -3, c= 1.

Hence, D = 9 - 4r = 0

                      r = \frac {9} {4} = \frac {a} {b}

Thus, r= 9+4= 13

We make the substitution 3^x=y, and multiplying both sides by y gives ry^2-3r+1=0. Since there is a unique real solution x to the original equation, there is thus a unique real solution y to our current equation, and the discriminant of the quadratic must be zero, so 9-4r = 0 --> r=9/4

Let y= 3^x

Then, 3^(-x) = 1/y.

The given equation now becomes ry + 1/y = 3

Or, ry^2 - 3y + 1 = 0

Then, the solutions for y are...

Root (i) :- {3 + sqrt(9-4r)}/2r Root (ii):- {3 - sqrt(9-4r)}/2r

Again, putting y= 3^x, we obtain that the solutions for x are...

Root (i):- [log{3 + sqrt(9-4r)}]/2r/log(3) Root (ii):- [log{3 + sqrt(9-4r)}/2r]/log(3)

For these roots to be equal, we must have...

9-4r= 0 Or, r = 4/9 Now comparing this with a and b we obtain a= 4 and b= 9.

So, a+b= 4+9= 13

Note that by the AM-GM inequality,

$r3^x+3^{-x} \geq 2 \sqrt{r}$

is equal if and only if $r3^x=3^{-x}$ .

This means $r = 3^{-2x}$ , which gives 1 solution for x.

Therefore, we must have $3 = 2 \sqrt{r}$ .

Thus, by solving that, we get $r = \frac{9}{4}$ .

Hence $a+b = 9+4 = 13$ .

we have the given equation is equivalent to $r3^x + \frac {1} {3^x} = 3$ or $rz + \frac {1}{z} = 3$ i.e $rz^2 + 1 - 3z = 0$ where $z=3^x$ We know the discriminant of the equation $rz^2 + 1 - 3z = 0$ is $\Delta = (-3)^2 + 4 \times r \times 1$ i.e $\Delta=9 - 4r$ . The equation only has a unique real solution so $\Delta = 0$ i.e $r = \frac {9}{4}$ i.e $\frac {a}{b} = \frac {9}{4}$ i.e $a + b = 9 + 4 =13$

Let y = 3^x. We can have this substitution since the function f: x --> 3^x is bijective, i.e, if f(a) = f(b), then a = b. The equation becomes ry + 1/y = 3. Since y can never be zero, we can multiply the equation with y, giving us with ry^2 - 3y + 1 = 0. This is quadratic and to achieve a unique solution in y, the discriminant 3^2 - 4r = 9 - 4r should be zero. Clearly, r = 9/4 and the required answer is 9 + 4 = 13.

Rewrite as ( r=3^{-x}(3-3^{-x}) )\ Let ( y=3^{-x} ). Then ( r=y(3-y) )\, but only for 1 such ( y ). Thus ( y = 3-y )\, and hence ( r = \frac{9}{4} )\

Let /(y=3^x}. Then $ry^2-3y+1=0$ . There exists a unique solution only if the discriminant is 0. Therefore, $9-4r=0$ and the answer is $13$ . (We can solve for the unique x by finding y then x).