Find x $$(3x-4)^{2x^2+2} =(3x-4)^{5x}$$
Iv'e proceeded by evaluating different cases.
- First case both bases = 1 and exponent is arbitrary $$\implies 3x-4=1$$ $$\therefore x=\frac{5}{3}$$
- Second case both bases = 0 and exponent is arbitrary $$\implies 3x-4=0$$ $$\therefore x=\frac{4}{3}$$
- Third and final is that both bases are equal so we solve for exponents $$\implies2x^{2}+2=5x$$ $$\therefore x= \frac{1}{2}$$or$$ x=2$$
Are these all the possible solutions? Am I missing something? Any insight would be of much help.
I assume we're dealing with real values alone throughout.
I would proceed as you do in stage (3) above, then substitute $x=2,\,1/2$ into $3x-4$ to see whether I get negative or positive values. Thus, if $x=2,$ we have that $3x-4=3(2)-4=6-4>0,$ so this is a solution. If $x=1/2$ we have that $3x-4=3(1/2)-4<0.$ Thus this cannot be a solution since $(-y^2)^{1/2}$ is not real-valued.
As for your other numbers, they work. Since there no other bases whose arbitrary powers always yield the same real value, then your set is complete.