Please Help me with this I think i figured out question 1... but I get no solution... please help me start number 2 or if you can show full solution that be sick thanks.
- $\log_{3x}(81)=2$
- $\log_3(\log_x(\log_4 16))=-1$
my attempt for number 1 ! :)
$$\frac{1}{81}\log_{3x}(81)=\frac{2}{81}$$ $$\log_{3x}=2/81$$ $$10^{\large{\log_{3x}}}=10(2/81)$$ $$3x=10 (2/81)$$ $$x=10 (2/81) /3 $$
I make multiple use of the rule: $\log_a x=b\implies a^b=x$
The first problem is already addressed in one of the other answers, but when you see $\log_{3x}(81)=2$, you should think $(3x)^2=9x^2=81\implies x=3$
As for the second problem, since $4^2=16$, we know that $\log_4 16=2$ and we can make the following simplification:
$$\log_3(\log_x(\log_4 16))=-1\\ \log_3(\log_x(2))=-1$$
Now, if we think of $\log_x(2)$ as a variable, call it $y$, we can rewrite the above equality like so: $$\log_3 y=-1\\ y=3^{-1}=\frac{1}{3}$$ Now that we have a value for $y$, lets put it back in context: $$y=\log_x(2)=\frac{1}{3}\\ x^{\frac{1}{3}}=2\\ x=8 $$
$$\log_{3x}(81)=2$$ $$(3x)^{\log_{3x}(81)}=(3x)^{2}$$ $$81=9x^2$$ $$9=x^2$$ $$3=x$$
