# Thread: How to solve for x??: (2^x) + (3^x) = 13

1. ## How to solve for x??: (2^x) + (3^x) = 13

How to solve for x??:

(2^x) + (3^x) = 13

2. ## Re: How to solve for x??: (2^x) + (3^x) = 13

$2^2= ?$, $3^2= ?$.

3. ## Re: How to solve for x??: (2^x) + (3^x) = 13

Originally Posted by HallsofIvy
$2^2= ?$, $3^2= ?$.
It is clear that x=2 but how do we arrive at this conclusion using algebraic method?

4. ## Re: How to solve for x??: (2^x) + (3^x) = 13

Originally Posted by MatthiasChampagne
It is clear that x=2 but how do we arrive at this conclusion using algebraic method?
You don't arrive at this conclusion using algebraic methods.

5. ## Re: How to solve for x??: (2^x) + (3^x) = 13

Can't we use logarithm here sirs??

6. ## Re: How to solve for x??: (2^x) + (3^x) = 13

Hi,
no, use logaritnm don't helps.
The problem haven't regulary algebric solution,

but simply x=1 or x=2 or x=3?
Probe, and see...

7. ## Re: How to solve for x??: (2^x) + (3^x) = 13

thanks sirs...so basically,it's just trial and error???

8. ## Re: How to solve for x??: (2^x) + (3^x) = 13

Knowing there will only be one root from the properties of exponential functions, we could use a root finding technique, such as Newton's method, which converges to $x=2$ in 10 iterations on my calculator with an initial guess of $x_0=0$.

I would use inspection first to see if a rational root exists, but say the equation had instead been:

$2^x+3^x=12$

Now, using Newton's method, we find:

$x\approx1.91768594489$

9. ## Re: How to solve for x??: (2^x) + (3^x) = 13

,.,thank you for that idea sir,.but i just got a little confused in using Newton's method,.,can you please elaborate a little on the steps sir??thnx

10. ## Re: How to solve for x??: (2^x) + (3^x) = 13

Have you studied differential calculus? Newton's method uses differentiation in its recursive approach. Newton's method essentially finds the x-intercept of the line tangent to the function for which we are trying to find a root at a value of x and uses that as a better approximation. Then this method is repeated recursively to converge to a root of the function. There are functions for which this doesn't always work for particular values of x, but for well-behaved functions like exponential functions, this is a good technique to use.

11. ## Re: How to solve for x??: (2^x) + (3^x) = 13

,.yes sir,.but when i substituted the values,.i got it all wrong.,.,i have the derivative (ln(3)*(3^x)) + (ln(2)*(2^x)) for the function,.but i got messed up for the substitution. help

12. ## Re: How to solve for x??: (2^x) + (3^x) = 13

Originally Posted by aldrincabrera
,.yes sir,.but when i substituted the values,.i got it all wrong.,.,i have the derivative (ln(3)*(3^x)) + (ln(2)*(2^x)) for the function,.but i got messed up for the substitution. help
If you know calculus, then looking at the derivative tells you that $2^t+3^t-13$ in everywhere increasing. It means that $t=2$ is the only root.

So no need for Newton's method.

13. ## Re: How to solve for x??: (2^x) + (3^x) = 13

Originally Posted by aldrincabrera
,.yes sir,.but when i substituted the values,.i got it all wrong.,.,i have the derivative (ln(3)*(3^x)) + (ln(2)*(2^x)) for the function,.but i got messed up for the substitution. help
You have the correct derivative.

For the function:

$f(x)=2^x+3^x-k=0$

$f'(x)=\ln(2)2^x+\ln(3)3^x$ and so by Newton's method, we have the recursion:

$x_{n+1}=x_n-\frac{2^{x_n}+3^{x_n}-k}{\ln(2)2^{x_n}+\ln(3)3^{x_n}}$

As stated, when $k=13$ there certainly is no need for this technique, but I simply wanted to give a method for approximating the root when it is irrational, since you seemed to be interested in an "algebraic" method.

14. ## Re: How to solve for x??: (2^x) + (3^x) = 13

You obviously didn't read the previous posts AND you have replaced the tacky link in your signature that you were asked to remove...nicely done!

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# SOLUTION TO 3^x 2^x=13

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