# How to solve an exponential function with multiple addends

Our math teacher gave us the following exponential equation to solve:

$3^x+10=2*7^x$

...and I was stumped. Eventually, the solution given was to graph both sides and find their intersection using a calculator. But that solution doesn't satisfy me.

Is it possible to solve this problem algebraically? A full solution is not necessary - I'd just like to know if it's possible, and maybe some pointers towards solving it. I anticipate that a proper solution will probably involve concepts outside of my grasp, but if so, that is okay.

I don't think there is a closed form solution to this equation.

Had it be $3^x +10=2.9^x=2.3^{2x}=2.(3^x)^2$ instead, you could have set $t=3^x$ to turn it to an algebraic equation, easily solved: $t+10=2.t^2$.

Trying the same trick here leads to $t+10=2.t^{\frac{log(7)}{log(3)}}$. The exponent ($1.7712437...$) is irrational so that there is no way of turning the equation into an algebraic one (a polynomial equation if you prefer).

This is not enough to say that the equation has no closed formula, but it is a good indication. The solution must be a transcendental number (http://en.wikipedia.org/wiki/Transcendental_number).

• I thought I was missing something simple, but apparently not. Thank you for your thoughts on this problem! – gengkev Mar 22 '14 at 0:18

Consider $f(x) = 3^x + 10 - 2*7^x$, and $f(0) = 9 > 0$, $f(1) = -1 < 0$, So by IVT, there is $0 < c < 1$ such that $f(c) = 0$. If $x < 0$, then $f(x) > 0$, and if $x > 1$, then $f'(x) = ln3*3^x - 2ln7*7^x < 0$, so $f(x) < f(1) = -1$. So there is no solution for $x < 0$ or $x > 1$. So you may use Newton's approximation method to find the solution. Start with say, $c = .5$

• Your ideas are marvellous. But please use latex. – Yiyuan Lee Mar 21 '14 at 7:29
• Newton's approximation method seems interesting. I will definitely take a look at that. – gengkev Mar 22 '14 at 0:20