# What is the solution? Factoring and computing the equation.

If you will be gracious enough to answer, the equation is currently: $$10^x + 15^{x-1}= 20,$$ What is the value of $2x^2$? Please list all steps, if you don't mind. To follow up, what is the name of the website that lists tags for mathematical signs on this website?

• – Zev Chonoles Feb 16 '14 at 9:15
• i give my thanks to you. – Diamond Louis XIV Feb 16 '14 at 9:21
• I do not think that there is an analytical solution. Are you looking for a numerical method ? – Claude Leibovici Feb 16 '14 at 9:55

I believe (unfortunately, without evidence) that the given equation is a transcendental equation, and am therefore resolving to employ the numerical Newton-Rhapson method, by finding the root(s) of $10^x + 15^{x-1} - 20 = 0$.
Set $$f(x) = 10^x + 15^{x-1} - 20$$ Then, $$f'(x)=10^x\ln10+15^{x-1}\ln15$$ Define the sequence $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Then the Newton-Rhapson method says that $x_n$ gives a root of $f(x) = 0$ for some sufficiently large $n$, with an initial guess $x_0$.
We take $x_0=1$ (a random guess), and find that the first few values of $x_n$ are: $$x_1 \approx 1.34932326879814\\ x_2 \approx 1.26510504583041\\ x_3 \approx 1.25547389505072\\ x_4 \approx 1.25535183049781\\ x_5 \approx 1.25535166963961\\ x_6 \approx 1.25535166945069\\ x_7 \approx 1.25535166945047\\ x_8 \approx 1.25535166945047\\ x_9 \approx 1.25535166945047$$ Thus, we see that $10^x + 15^{x-1} - 20 = 0$, or $10^x + 15^{x-1} = 20$ when $x\approx1.25535166945047$.
• Hmm...I think you're confusing your exponent laws. In general, $a^{bc} = (a^b)^c$ and $a^{b+c} = a^ba^c$. This means that: $$10^{x \log_{10} 15} \neq 10^{\log_{10} 15}10^x$$ – Adriano Feb 16 '14 at 9:41
• @Yiyuan Lee. If $x=1$ is a solution, I must then admit that $11=20$ ! Numerically, the solution seems to be $x=1.25535$ and $2x^2$ looks to be very close to $\pi$. – Claude Leibovici Feb 16 '14 at 9:42