## anonymous one year ago 3log(x-1)=4sinx on [0, 3pi)

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1. anonymous

i have to solve for x

2. anonymous

Using a numerical approach, presumably? You can use the Newton-Raphson method. Define $$f(x)=3\log(x-1)-4\sin x$$, where I'm assuming $$\log$$ denotes the natural logarithm. Just to sample some values of $$f$$, let's consider $$x=2$$ and $$x=\pi\approx3.14$$. $\begin{array}{c|c} \text{sample point }x^*&f(x^*)\\ \hline 2&3\log(2-1)-4\sin2=-4\sin2<0\\ \pi&3\log(\pi-1)-4\sin\pi=3\log(\pi-1)>0 \end{array}$ since $$\sin x>0$$ for $$0<x<\pi$$, and $$3\log(x-1)>0$$ for all $$x>2$$ (including $$\pi$$). Therefore, by the intermediate value theorem there's a value $$2<c<3$$ such that $$f(c)=0$$ - this is the root you want. Suppose $$x_0=2.5$$, just as a convenient starting point. Use the N-R formula to approximate $$c$$ to within whatever accuracy you desire: $f(x)=3\log(x-1)-4\sin x\implies f'(x)=\frac{3}{x-1}-4\cos x\\[2ex] \begin{cases}x_0=2.5\\[1ex]x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)},&n\ge1\end{cases}$As $$n\to\infty$$, you have $$x_n\to c$$.