# x=(1.5)^x

• April 4th 2011, 12:16 AM
Riddler
x=(1.5)^x
Hi,
I don't remember how can I solve it for example: x=1,5^x
The Riddler
• April 4th 2011, 12:24 AM
Prove It
I assume this equation is $\displaystyle x = 1.5^x$. You can't solve it exactly, you have to use a numerical method.
• April 4th 2011, 12:39 AM
earboth
Quote:

Originally Posted by Riddler
Hi,
I don't remember how can I solve it for example: x=1,5^x
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.
• April 4th 2011, 03:03 AM
HallsofIvy
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 $f(x)= xe^x$.

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.
• April 4th 2011, 04:19 AM
Riddler
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?
• April 4th 2011, 05:16 AM
e^(i*pi)
It also has a solution of $x \approx 1.18$

Not sure how you'd show it though, I used wolfram
• April 4th 2011, 05:54 AM
Prove It
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...
• April 4th 2011, 09:13 AM
Quacky
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.
• April 5th 2011, 04:47 AM
Riddler
I wrote a program yesterday, which has solved my problem.