What is log_{sqrt(2)}(log_{sqrt(2)}(16)) ?

The solution to log_{sqrt(2)}(log_{sqrt(2)}(16)) is 6

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simplify log_{sqrt(2)}(log_{sqrt(2)}(16))

simplify

lo g_{2} (lo g_{2} (16))

6

Solution steps

lo g_{2} (lo g_{2} (16))

Apply radical rule:

a = a^{\frac{1}{2}}

2 = 2^{\frac{1}{2}}

= lo g_{2^{\frac{1}{2}}} (lo g_{2} (16))

Apply log rule:

lo g_{a^{b}} (x) = \frac{1}{b} lo g_{a} (x), a > 0

lo g_{2^{\frac{1}{2}}} (lo g_{2} (16)) = \frac{1}{\frac{1}{2}} lo g_{2} (lo g_{2} (16))

= \frac{1}{\frac{1}{2}} lo g_{2} (lo g_{2} (16))

lo g_{2} (16) = 8

= \frac{1}{\frac{1}{2}} lo g_{2} (8)

Factor the number:

8 = 2^{3}

= \frac{1}{\frac{1}{2}} lo g_{2} (2^{3})

Apply log rule:

lo g_{a} (a^{b}) = b

lo g_{2} (2^{3}) = 3

= \frac{1}{\frac{1}{2}} \cdot 3

\frac{1}{\frac{1}{2}} \cdot 3 = 6

= 6

e^{x} \cdot (- e^{- x})

(a + 3) (a - 1)

(3 c - 5)^{2}

3 \div 1

2 \times \frac{1}{3}

What is log_{sqrt(2)}(log_{sqrt(2)}(16)) ?
The solution to log_{sqrt(2)}(log_{sqrt(2)}(16)) is 6