Why is $ \frac12 \cdot (\ln 4)^2 = 2 \cdot(\ln 2)^2 $?

Question

Why are these equal? Can somebody explain?

$$ \frac12 \cdot (\ln 4)^2 = 2 \cdot(\ln 2)^2 $$

Answer 1

$$\frac { 1 }{ 2 } \cdot (\ln { 4 } )^{ 2 }=\frac { 1 }{ 2 } { \left( \ln { { 2 }^{ 2 } } \right) }^{ 2 }=\frac { 1 }{ 2 } { \left( 2\ln { 2 } \right) }^{ 2 }=\\ =\frac { 1 }{ 2 } { 2 }^{ 2 }{ \left( \ln { 2 } \right) }^{ 2 }=2{ \left( \ln { 2 } \right) }^{ 2 }$$

Answer 2

See that

$$\frac{1}{2}(\log_e4)^2$$

$$=\frac12\log_e4\times\log_e4$$

$$=\frac12\log_e2^2\times\log_e2^2$$

$$\frac12\times4\log_e2\times\log_e2=2(\log_e2)^2$$

Theorem used-$\log_e a^m=m\log_e a$

Note that $\log_e(x)=\ln(x)$

Answer 3

Fondamental property of logarithms:

$$\log(a^n) = n\log(a)$$

In which it must be $a > 0$.

So what you have is simply

$$\frac{1}{2}\left(\log 4 \cdot \log 4\right) $$

But now we can see $4$ as $2^2$ hence

$$\frac{1}{2}\left(\log 2^2 \cdot \log 2^2\right) $$

Now for what I told you at the beginning, you can write

$$\log 2^2 = 2\log 2$$ thence

$$\frac{1}{2}\left(2\log 2 \cdot 2\log 2\right) $$

Which is equal to

$$\frac{4}{2}\left(\log 2 \cdot \log 2\right) = 2(\log 2)^2$$