Tasty Logarithm #2

We will require that both $\sin(x)$ and $\cos(x)$ are positive, i.e., that $x$ is in the first quadrant. This in turn implies that $\sin(x) + \cos(x) \gt 0$ as well.

From the first equation we have that $\sin(x)\cos(x) = 10^{-1} = \dfrac{1}{10}.$

From the second equation we have that

$\log_{10}(\sin(x) + \cos(x)) = \log_{10}(\sqrt{n}) - \dfrac{1}{2} =$

$\log_{10}(\sqrt{n}) - \log_{10}(\sqrt{10}) = \log_{10}\left(\sqrt{\dfrac{n}{10}}\right).$

Thus $\sin(x) + \cos(x) = \sqrt{\dfrac{n}{10}} \Longrightarrow (\sin(x) + \cos(x))^{2} = \dfrac{n}{10}$

$\Longrightarrow 1 + 2\sin(x)\cos(x) = \dfrac{n}{10}.$

Now substitute in the result from the first equation to see that

$1 + 2*\dfrac{1}{10} = \dfrac{n}{10} \Longrightarrow \dfrac{12}{10} = \dfrac{n}{10} \Longrightarrow n = \boxed{12}.$

All logs in this solution are to base ten.

First note that from the first equation

$\log(\sin x\cos x)=-1$

$\implies \sin x\cos x=\frac{1}{10} \dots (1)$

Solve the second equation to get

$\log(n)=2\log(\sin x+\cos x)+1=\log((\sin x+\cos x)^2) +\log(10)$

$\implies \log n=\log(10(1+2\sin x\cos x))$

$\implies n=10(1+2\sin x\cos x)$

Substituting the value from (1) gives

$n=10(1+0.2)=\boxed{12}$

Obviously, sinxcosx=1/10 . Multiplying 2 to both sides, 2sinxcosx=1/5 Then, adding 1 to both sides, we have 1+2sinxcosx=6/5 The LHS can be factored out as (sinx+cosx)^2=6/5 ---->(equation1)

The second equation is equivalent to log(sinx+cosx)^2=logn-log10 (Note if the logarithm has no base, it is understood that the base is 10) Plugging in our value in equation 1, we have log(6/5)+log10=logn Thus, solving for n yields 12