# Thread: a^x = b^x + c

1. ## a^x = b^x + c

how does one solve an equation of the type a^x = b^x + c??

for the life of me i cant do it.

2. Originally Posted by ayoyoayoyo
how does one solve an equation of the type a^x = b^x + c??

for the life of me i cant do it.
To get you started, take the natural log of both sides of the equation.

3. tried that but the constant term got in the way

4. Originally Posted by ayoyoayoyo
tried that but the constant term got in the way
Isn't it true that ln (ax) = ln(a) + ln(x)?

5. i can not see how that pertains to the question at hand

we have ln(b^x+c)

6. Originally Posted by wonderboy1953
Isn't it true that ln (ax) = ln(a) + ln(x)?
I'll go further:

ln(a^x) = ln(b^x +c)= ln(b^x)ln(c), then

xln(a) = xln(b)ln(c); ln(c) = xln(a)/xln(b) = ln(a)/ln(b) = ln(a) - ln(b)

7. The logarithm of a sum is not something you can simplify. The identity only works the other way.

With this problem, I'd probably go for a numerical solution using Newton-Raphson. Of course, this assumes you know a, b, and c, and that a and b are both positive. Is that the case?

8. so no analytic solution?

9. None of which I am aware. Both Mathematica and WolframAlpha fail to give an analytical solution. I should point out that not every combination of a, b, and c will admit a solution, even if all of them are positive.

10. how do you use newton's to find the solution? isnt newtons only used to find the roots of a f()?

11. Right, but you can always convert an equation-solving problem into a root-finding problem by throwing everything on to one side of the equation thus:

$\displaystyle f(x)=a^{x}-b^{x}-c=0.$

Then you use Newton-Raphson on $\displaystyle f(x).$

If you have convergence problems, and your solutions are oscillating wildly, then you probably don't have a solution for that particular combination of a, b, and c. That's a little warning sign you can look for.