Natural log and trigonometry

Algebra Level 3

Consider the following equations

$\begin{aligned} \ln(\sin x)+\ln(\cos x)&=-1\\ \ln(\sin x+\cos x)&=\frac{1}{2}(\ln(y+e)-1) \end{aligned}$

Find the value of $y$ .

The answer is 2.

1 solution

Sravanth C.
May 28, 2016

Relevant wiki: Logarithms

$\begin{aligned} \ln(\sin x) + \ln(\cos x) &= -1\\ \ln(\sin x\cos x) &= -1\\ \color{#D61F06}{\sin x\cos x} &= \color{#D61F06}{e^{-1}} \end{aligned}$

Now, we can simplify the second expression as follows;

$\begin{aligned} \ln(\sin x+\cos x)&=\frac{1}{2}(\ln(y+e)-1)\\ \ln(\sin x +\cos x)^2&=\ln(y+e)-\ln e\\ \ln(\sin^2 x+\cos^2 x + 2\color{#D61F06}{\sin x\cos x}) &=\ln\left(\dfrac{y+e}{e}\right)\\ 1 + 2\color{#D61F06}{e^{-1}}&=\dfrac{y+e}{e}\\ y+e&=e+2\\ y&=\boxed 2 \end{aligned}$

Moderator note:

Note that you should verify that such a value of $x$ exists.

Even though your got the answer but your solution is not $100\%$ correct.

First of all, you didn't calculate the domain of $y$ which made it compulsory for you to check the obtained solution ,merely, by plugging it in the given equation(s).

Note that, these small things might appear trivial, but they are of immense importance.

Aditya Sky - 5 years ago

Thanks you very much! Short and clear solution!

Tommy Li - 5 years ago

It should be:

$\begin{aligned} 1 + 2e^{-1} & = \frac{y\color{#D61F06}{+}e}{e} \\ 1 + \frac{2}{e} & = \frac{y}{e} + 1 \\ \implies y & = 2 \end{aligned}$

Chew-Seong Cheong - 5 years ago

Oops, thanks sir. Edited.

Sravanth C. - 5 years ago

Natural log and trigonometry

The answer is 2.

1 solution

Moderator note:

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