# Is this equation system solvable?

As the title states: is this equation system solvable?

$$x+y = 3 \\ 2^x + 3^x = 45$$

And by solvable I mean doing it using pen and paper, no computing the result or approximations.

• Shouldn't one of the exponents in the second equation be $y$?
– John
Commented Jan 10, 2014 at 17:58
• Shouldn't there be $35$ instead of $45$ ? :-) Commented Jan 10, 2014 at 21:52

Obviously $y = 3-x$. The difficult part is the other equation. We can write it as: $$45=2^x+3^{3-x} = 2^x+\frac{27}{3^x}$$ or: $$3^x(45-2^x) - 27 = 0$$ Let's denote the left hand side by $f(x)$. We have: $$\lim_{x\to-\infty}f(x)=-27$$ $$\lim_{x\to\infty}f(x)=-\infty$$ $$f(0)=17$$ Thus $f(x)=0$ has at least two solutions. However, I wouldn't know hot to find them if not numerically.

we can go further for the solution:

case 1:$x<0, 2^x<1 ,45-2^x>44$

$27=3^3,3^4=81>45=5 \times 3^2 >3^3, \implies -2<x<0, f(-1)=-22,f(-.5)=-1.4,f(-.25)=6.55$

so we can sure $x$ close to $-.5$, may be $-0.4$ mostly.

case 2:$x>0, 2^6=64>45, 2^5=32 < 45, f(5)=3132,f(6)=-13878,f(5.5)=-107,$ now we know the answer is close and less $5.5$.

rest need calculator. $2^{5.4}<45,f(5.4)=1019,f(5.45)=485,f(5.475)=187,f(5.49)=-3$ we can see the answer is very close to $5.49.(5.489771)$