Write a simpler expression for $y = 4\ln\!\left(\sin 4t\right)-\ln\!\left(\frac{\sin^{4} 4t}{7}\right)$

Question

Write a simpler expression for $y = 4\ln\!\left(\sin 4t \right)-\ln\!\left(\frac{\sin^{4} 4t }{7}\right) \;\;$ which does not involve the variable t.

So, uhm, how do I even start?

Sure, I can rewrite the second term as:

$-4\ln \left(\frac{\sin 4t}{7}\right)$

But how do I get rid of t?

Answer 1

\begin{align} y & = 4\ln\big(\sin 4t \big) - \ln\left(\frac{\sin^{4} 4t }{7}\right) \\ & = 4\ln\big(\sin 4t \big) - \Big( \ln\big(\sin^4 4t \big) - \ln 7\Big)\\ & = 4\ln\big(\sin 4t \big) - \ln\big(\sin^4 4t \big) + \ln 7\\ & = 4\ln\big(\sin 4t \big) - 4\ln\big(\sin 4t \big) + \ln 7\\ & = \ln 7 \end{align}

Answer 2

Notice that:

$y = 4\ln\!\left(\sin\!\left(4t\right)\right)-\ln\!\left(\frac{\sin^{4}\!\left(4t\right)}{7}\right)= 4\ln\!\left(\sin\!\left(4t\right)\right)-(4\ln\!\left(\sin\!\left(4t\right)\right)-\ln(7))=\ln7$

Attention: $\ln\dfrac{\sin^4(4t)}{7}\neq4\ln\dfrac{\sin(4t)}{7}$

Answer 3

The full answer to this is that it can't be done if you properly account for domains. At $t=1$ this expression is undefined. At $t=2$ it is defined. So the expression certainly depends somehow on $t$.

If $\sin(4t)$ is zero or negative, the entire expression is undefined. If $\sin(4t)$ is positive, then logarithm rules (which you have not applied correctly because the exponent of $4$ does not apply to the denominator of $7$) lead to it being equal to $\ln(7)$.

So the expression is kind of periodic in $t$. For an interval of length $\pi/4$ it is constant, and for the next interval of the same length it is undefined. There's no way to express such an expression without using an input variable somehow. Either directly, as in the original expression, or by stating that the input variable can only come from a certain restricted domain.

The problem author likely overlooked this, and expects a constant answer of $\ln(7)$. They could achieve this by assuring the argument to $\ln$ is positive, say by giving you $$4\ln(\sin(4t)+3)-\ln\mathopen{}\left(\frac{(\sin(4t)+3)^4}{7}\right)\mathclose{}$$

Answer 4

Hint:

Use the rules $$\ln(A^B) = B\ln(A)$$ and $$\ln(C/D) = \ln(C) - \ln(D).$$