<s>[INST] Solving $\frac{3^x+2^x}{3^x-2^x}=7$: More than one answer? How to solve for all $x$? 
I am trying to solve this problem (math for fun):
$$\frac{3^x+2^x}{3^x-2^x}=7$$

Step 1. Let$\:a=3^x\:and\:b=2^x$
Step 2. $\frac{a+b}{a-b}=7$
Step 3. $a+b=7\left(a-b\right)=7a-7b$
Step 4. $6a-8b=0$
Step 5. $6a=8b=3\cdot 2\cdot a=4\cdot 2\cdot b$
Step 6. $3a=4b$
Step 7. $3\cdot 3^x=2\cdot 2\cdot 2^x=3^{x+1}=2^{x+2}$
Step 8. Log both sides
Step 9. $\ln \left(3\right)+x\ln 3=x\ln2+\ln 4$
Step 10. $\ln \left(3\right)-\ln 4=x\:\left(\ln 2-\ln 3\right)$
Step 11. $\frac{\:\left(\ln 3-\ln 4\right)}{\left(\ln 2-\ln 3\right)\:}=x≈0.71$
SOLVED^
 [/INST] $$3^x+2^x=7\cdot3^x-7\cdot2^x$$
$$8\cdot2^x=6\cdot3^x$$
$$\left(\frac32\right)^x=\frac 43$$
and
$$x=\log_{3/2}\frac43.$$
As every steps are equivalent, this is the only solution.
</s>