x=(1.5)^x

**Riddler** · April 4th 2011, 01:16 AM

Hi,
I don't remember how can I solve it for example: x=1,5^x
Help me please,
The Riddler

**Prove It** · April 4th 2011, 01:24 AM

I assume this equation is $\displaystyle x = 1.5^x$ . You can't solve it exactly, you have to use a numerical method.

**earboth** · April 4th 2011, 01:39 AM

Originally Posted by Riddler

Hi,
I don't remember how can I solve it for example: x=1,5^x
Help me please,
The Riddler

This equation doesn't have a real solution because $(1.5)^x > x\ ,\ x \in \mathbb{R}$

I've checked at which x-value the function $f(x)=(1.5)^x$ has the slope 1. I got the point

$P\left(\dfrac{\ln\left( \frac1{\ln(1.5)} \right)}{\ln(1.5)}~,~\dfrac1{\ln(1.5)} \right)$

and this point lies above the line y = x.

**HallsofIvy** · April 4th 2011, 04:03 AM

I will admit that I would have interpreted the question as "if x= 1, what is $5^x$ ", but solving the equation $x= 1.5^x$ is much more interesting.

Of course, $1.5^x= e^{ln(1.5^x)= e^{xln(1.5)}$ so $x= 1.5^x$ is the same as $x= e^{xln(1.5)}$ and that is the same as $xe^{x(-ln(1.5))}= 1$ . Let y -ln(1.5)x so that $x= \frac{y}{-ln(1.5)}$ and the equation becomes $\frac{y}{-ln(1.5)}e^y= 1$ or $ye^y}= -ln(1.5)$ .

Now, we have y= W(-ln(1.5)) where W is "Lambert's W function", the inverse function to [itex]f(x)= xe^x[/itex].

Finally, then $x= \frac{y}{-ln(1.5)}= \frac{W(-ln(1.5))}{-ln(1.5)}$ .

Of course, as earboth says, x is not a real number because the domain of the real valued W function, while it includes some negative numbers, does not include any that negative.

**Riddler** · April 4th 2011, 05:19 AM

My previous example was wrong.
But if: x=(1.15)^x
It has got a real solution: x=22.1723..
How can I solve it?

**e^(i*pi)** · April 4th 2011, 06:16 AM

It also has a solution of $x \approx 1.18$

Not sure how you'd show it though, I used wolfram

**Prove It** · April 4th 2011, 06:54 AM

Drawing the graphs of $\displaystyle y = x$ and $\displaystyle y = 1.15^x$ and noting where they intersect gives you an idea of the solutions...

**Quacky** · April 4th 2011, 10:13 AM

Of course, you could use any number of methods of approximation. I tried the risky approach of blind iteration, using the following steps (having drawn the graphs first for a rough idea of the roots, obviously):
$x_{n+1}=1.15^{x_n}$

Starting at $x_1=20$

$x_2=1.15^{20}=16.36$

$x_3=1.15^{16.36}=9.85$

$x_4=1.15^{9.85}=3.96$
....
We get the $x\approx~1.18$ solution at approximately the 9th iteration. However I was trying to get the other solution! You could try a slightly better approach, such as interval bisection, Newton-Raphson or linear interpolation or something, but then you probably know all this already.

**Riddler** · April 5th 2011, 05:47 AM

I wrote a program yesterday, which has solved my problem.
You can download it here: http://strategist.fw.hu/program.rar

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