Recent content by wit

Been off of math for quite some time now. It seems like even the most basic rules managed to escape my mind. I cannot be thankful enough for your help. :) Take care!

$$\log_3(x)=-2$$ | $$\log_a(x)=b => a^{b} = x$$ $$3^{-2}=x$$ $$x=\frac{1}{9}$$ and as you said $x=1$ Final answer: {$$1,\frac{1}{9}$$} ... How do I get to $$x=1[/math], though? It's pretty straight forward that $$1^{anything} = 1$$ but how do I 'prove' it?

$$(3x)^{\log_3(x)}=\frac{1}{x}$$ $$(3)^{\log_3(x)}*(x)^{\log_3(x)}=\frac{1}{x}$$ So now I can apply: $$(a)^{\log_a(b)} = b$$? Which means: $$(x)*(x)^{\log_3(x)}=\frac{1}{x}$$ | /x $$(x)^{\log_3(x)}=\frac{1}{x^2}$$ And now I think I am confused

Thank you for your reply. Tried my best, is it any good? $$(3x)*(3x)^{\log_3(x)}=3$$ $$(3x)*(x^2)=3$$ (Not sure about this part) $$(3x^3)=3$$ $$(x^3)=1$$ $$(x)=1$$

$$(3x)^{1+\log_3(x)}=3$$ How do I solve this one?