What properties of exponents and logarithms is used here?

Question

In a solution to one of my assignment is given that:

$$4^{log_{16} n} = n^{log_{16} 4} = n^{\frac12}$$

I understand the 2nd part, that's simple, but how was the first part achieved?

FYI, I stumbled on the identity $a^{\log_b c} = c^{\log_b a}$ recently --- see my comments, and see "We now make use of the identity ..." in my answer, both at Tricky logarithmic question. See also Name for a Logarithm Identity/Property. — Dave L. Renfro, Oct 01 '18 at 14:43

score 1 · Answer 1 · answered Oct 01 '18 at 14:12

1

Indeed an important part is missing:

$4^{log_{16}n}= 16^{log_{16}4*log_{16}n} = n^{log_{16}4} $

answered Oct 01 '18 at 14:12

ultrahamster

score 0 · Accepted Answer · answered Oct 01 '18 at 14:16

0

Let $$y=4^{\log_{16}x}$$ $$\log_{16}y=\log_{16}(x^{\log_{16}4})$$ $$\log_{16}y=\log_{16}(x^{1/2})$$ $$y=x^{1/2}$$

answered Oct 01 '18 at 14:16

Tom Himler

2 Answers2