# X*a^x=b, how to solve this?

1. May 1, 2006

### raul_l

x*a^x=b

how to solve this?

2. May 1, 2006

### Curious3141

You can't, in general, solve it exactly. Use a numerical method to approximate the roots.

3. May 1, 2006

### VietDao29

This is a good candidate for Newton's method.
You first choose an x0 arbitrarily. You can graph it first, then choose an x0 wisely so it's near the roots.
Then use the formula:
$$x_{n + 1} = x_n - \frac{f(x_n)}{f'(x_n)}$$, and let n increase without bound to obtain the solution, i.e:
$$x = \lim_{n \rightarrow \infty} x_n$$.
Can you get this? :)

4. May 1, 2006

### raul_l

Yes.
And it works! :)
I didn't know about Newton's method before.

5. May 3, 2006

### Emieno

with math reasoning

Suppose a and b are unknown constants
Let $$y=xa^x$$ and hence $$y=b$$
Take a ln of bothe sides leading to $$lny=lnx+xlna$$
We understand that x,a,b must be > 0

Taking a derivative of y gives us
$$\frac{dy}{dx}=(\frac{1}{x}+lna)xa^x$$
Now we find that $$x=0 (omitted), \frac{-1}{lna}$$

Next, we draw a table to check signs of $$\frac{dy}{dx}$$, but before that we check $$a$$
1. if 0<a<1
Look at the table and mark for sign (+/-), then check for y to compare with $$y=b (a.straight.line)$$, which means you need to reason the value of b for where the root(s) exist.
2. if a>1
Do the same to find out root domain

Now things become easier when you know concrete constant a, b. just put them inthere to find a root. This way looks crary though but sovable domain can be understood

Last edited: May 3, 2006