Find the value of $x\;9^{x}+15^{x}=25^{x},\;x$ is real

Question

I tried: $a=3^{x}$ and $b=5^{x}$ $a^{2}+ab=b^{2}\Rightarrow ab=b^{2}-a^{2}=(b+a)(b-a)$, but I didn't get an answer.

If $a^2 + 5^x a -25^x = 0$ then $a =\frac {-5^x\pm \sqrt{5^{2x}+4*25^x}}2$ which can be simplified.... — fleablood, Apr 23 '20 at 00:43
Rewrite it as
$$\left(\frac{9}{25}\right)^x+\left(\frac{15}{25}\right)^x=1 \implies\left(\frac{3}{5}\right)^{2x}+\left(\frac{3}{5}\right)^x-1=0$$

Let $y=\left(\frac{3}{5}\right)^x$. Solve the quadratic in $y$, then substitute $x = \frac{\ln y}{\ln(\frac35)}$. — Axion004, Apr 23 '20 at 00:52
https://math.stackexchange.com/questions/1356236/what-is-the-solution-to-the-equation-9x-6x-2-cdot-4x-0 — lab bhattacharjee, Apr 23 '20 at 02:16

Quanto · Answer 1 · 2020-04-23T00:41:28.037

3

Let $y=(\frac35)^x$. Then, divide the equation by $15^x$ to get

$$y-\frac 1y+1=0$$

Solve to get

$$y = \frac{-1\pm\sqrt5}2$$

Use $x\ln\frac35 =\ln y$ to obtain the sulution

$$x= \frac{ \ln \frac{\sqrt5-1}2}{\ln\frac35}$$

edited Apr 23 '20 at 00:41

answered Apr 23 '20 at 00:25

Quanto

97,352

how do you get this? – Kayan Tchian Apr 23 '20 at 00:33
oh, ich verstehe jetzt – Kayan Tchian Apr 23 '20 at 00:36

score 2 · Answer 2 · answered Apr 23 '20 at 00:39

2

If $a^2 + ab = b^2$ then $a^2 + ab -b^2 =0$ and you can solve for $a$ in terms of $b$ or $b$ in terms of $a$.

$a = \frac {-b \pm\sqrt{b^2 +4b^2}}2=\frac {(-1\pm \sqrt 5)b}2$

so

$3^x = 5^x\frac {(-1\pm \sqrt 5)}2$

$(\frac 35)^x = \frac {(-1\pm \sqrt 5)}2$. As $(\frac 35)^x > 0$ we have

$x = \log_{\frac 35} \frac {(-1+ \sqrt 5)}2$

answered Apr 23 '20 at 00:39

fleablood

124,253

I find this answer so amazing compared to other answers. It's always better to write one $\log$ instead of two $\ln$s divided by each other. – Jun 07 '21 at 08:50
Well... to be fair $\frac 35$ is a non standard base and and many mathematicians prefer to use only the natural log and we prefer $\frac {\ln M}{\ln b}$ rather than $\log_b M$. And that's a fair preference. However I dont see any value to stating $\log_{whatevever} (\frac {-1+\sqrt 5}2)$ as $\log_{whatever} (-1+\sqrt 5) -\log_{whatever} 2$. – fleablood Jun 07 '21 at 16:18
I also prefer $\log_{b}(\frac{a}{c})$ rather than $\log_{b}(a)-\log_{b}(c)$; because the former is simpler. I prefer $\log_{b}(a)$ rather than $\frac{\ln{a}}{\ln{b}}$ for the same reason. btw, can you please explain why that's a fair preference? – Jun 08 '21 at 17:19
Because $\ln$ is a standard and universally accepted function and accepted notation for a value. So it is convenient and easy to simply convert anything to $\ln$ notation. Just as we are likely to express numbers numbers in base 10 rather than arbitrary bases like $7$ or $15$ arbitrarily. An experienced mathematician won't even think of $\frac {\ln a}{\ln b}$ as "the natural log of $a$ divided by the natural log of $b$" but is more likely to read it as "logarithm of $a$ based $b$ expressed in terms of natural logs". – fleablood Jun 08 '21 at 17:30

score 1 · Accepted Answer · answered Apr 23 '20 at 00:36

$$\begin{align}9^x+15^x&=25^x\\\left(\dfrac35\right)^{2x}+\left(\dfrac35\right)^x&=1\\y^2+y&=1\qquad\boxed{\text{Let }y=\left(\dfrac35\right)^x}\\y^2+y-1&=0\\y&=\dfrac{-1\pm\sqrt{5}}2\end{align}$$

$$\begin{equation}\begin{split}\left(\dfrac35\right)^x&=\dfrac{-1+\sqrt5}2\\x\ln\left(\dfrac35\right)&=\ln\left(\dfrac{-1+\sqrt5}2\right)\\x&=\dfrac{\ln(-1+\sqrt5)-\ln2}{\ln3-\ln5}\approx 0.942028\end{split}\qquad\begin{split}\left(\dfrac35\right)^x&=\dfrac{-1-\sqrt5}2\\x\ln\left(\dfrac35\right)&=\ln\left(\dfrac{-1-\sqrt5}2\right)\\x&=\dfrac{\ln(1+\sqrt5)-\ln2+i\pi}{\ln 3-\ln5}\in\mathbb{C}\end{split}\end{equation}$$

Find the value of $x\;9^{x}+15^{x}=25^{x},\;x$ is real

3 Answers3