This equation cannot be solved in terms of elementary functions. You can either use numeric methods like Newton's, or solve in terms of Lambert W function:
\begin{align*}
e^x - 20x &= 0 \\
20x &= e^x \\
xe^{-x} &= \frac{1}{20} \\
-x e^{-x} &= -\frac{1}{20} \\
-x &= W\left(-\frac{1}{20}\right) \\
x &= -W\left(-\frac{1}{20}\right)
\end{align*}
$$\cos^3x-\sin^3x=(\cos x-\sin x)(1+\sin x\cos x)$$
If we set $\displaystyle\cos x-\sin x=u, u^2=1-2\sin x\cos x\implies (\cos x-\sin x)(1+\sin x\cos x)=u\left(1+\frac{1-u^2}2\right)=\frac{3u-u^3}2$
So, the problem reduces to $\displaystyle u^3-3u+2=0$
Clearly, $1$ is a root and $\displaystyle\displaystyle\frac{u^3-3u+2}{u-1}=u^2+u-2=(u+2)(u-1)$
$\displaystyle\implies u^3-3u+2=(u+2)(u-1)^2$
If $\displaystyle u+2=0, u=-2,\cos x-\sin x=-2,$ $\displaystyle\cos x-\sin x=\sqrt2\cos\left(x+\frac\pi4\right)$
$\displaystyle\implies-\sqrt2\le\cos x-\sin x\le\sqrt2$
So, $u=\cos x-\sin x$ must be $\displaystyle=1\implies \cos x-\sin x=1\iff \sqrt2\cos\left(x+\frac\pi4\right)=1$
$\displaystyle\implies\cos\left(x+\frac\pi4\right)=\frac1{\sqrt2}=\cos\frac\pi4$
$\displaystyle\implies x+\frac\pi4=2m\pi\pm \frac\pi4$ where $m$ is any integer
Best Answer
Your equation is equivalent to $$\left(\frac{5}{3}\right)^x = \frac{9}{5}$$ Therefore, $x = \text{log}_{\frac{5}{3}} \frac{9}{5}$.