Correct one: Solve for x, #log_2 (log_3x) = 2#?

1 Answer
Jan 18, 2018

Answer:

I got the same answer as the other answerer, using a different method: #x=64#

Explanation:

Start with #log_2(log_3x)=2#

Raise both sides to the power of 2:

#(log_2(log_3x))^2=2^2=4#

Note that #log_2("anything")^2#= #"anything"#

That is, #log_2(3)^2=3#, #log_2(6)^2=6# and so on.

So then #log_3(x)=4#

Now raise both sides to the power 3:

#log_3(x)^3=4^3=64#

Same as above, #log_3(x)^3=x#

So #x=64#.