1. ## Equation problem

Hello!

Solve this equation for x: $\sqrt3^x+1=2^x$

Obviously, the solution is $x=2$. This can be proven by graphing $\sqrt3^x+1$ and $2^x$.

But I can't find any proper ALGEBRAIC way to solve this equation. Any idea?

Thank you.

2. Anyone?

3. Try to explore the function f(x)=(sqrt3)^x+1-2^x

4. See here for a related thread.

5. Maybe iteration?

$2^x = \sqrt{3} ^x + 1$

$x_{n+1} = \dfrac{ln(\sqrt{3}^{x_n} + 1)}{ln2}$

Take a positive value as a start. Then, take the result you get in the same formula again and again until you get consistent results.

For example, take the first as x = 1.5

$x_{1} = \dfrac{ln(\sqrt{3}^{1.5} + 1)}{ln2} = 1.713478979...$

Use this again,

$x_{2} = \dfrac{ln(\sqrt{3}^{1.713478979...} + 1)}{ln2} = 1.833140309...$

Again and again, until you get 1.999 which is enough to say that x = 2

EDIT: There was a LaTeX error which took me some time to spot... and before I knew it, Ackbeet posted the link...

6. Thank you very much!