Find the value of $x\;9^{x}+15^{x}=25^{x},\;x$ is real
I tried:$a=3^{x}$ and $b=5^{x}$$a^{2}+ab=b^{2}\Rightarrow ab=b^{2}-a^{2}=(b+a)(b-a)$, but I didn't get an answer.
$\endgroup$ 53 Answers
$\begingroup$If $a^2 + ab = b^2$ then $a^2 + ab -b^2 =0$ and you can solve for $a$ in terms of $b$ or $b$ in terms of $a$.
$a = \frac {-b \pm\sqrt{b^2 +4b^2}}2=\frac {(-1\pm \sqrt 5)b}2$
so
$3^x = 5^x\frac {(-1\pm \sqrt 5)}2$
$(\frac 35)^x = \frac {(-1\pm \sqrt 5)}2$. As $(\frac 35)^x > 0$ we have
$x = \log_{\frac 35} \frac {(-1+ \sqrt 5)}2$
$\endgroup$ 4 $\begingroup$Let $y=(\frac35)^x$. Then, divide the equation by $15^x$ to get
$$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}$$
$\endgroup$ 2 $\begingroup$$$\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}$$
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