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Question Number 83202 by Rio Michael last updated on 28/Feb/20

$$\mathrm{find}\mathrm{the}\mathrm{first}4\mathrm{terms}\mathrm{in}\mathrm{the}\mathrm{maclaurin}[$$$$\mathrm{series}\mathrm{expansion}\mathrm{for}\ln (1+3x)\mathrm{hence}\mathrm{show}\mathrm{that}$$$$\mathrm{if}{x}^2\mathrm{and}\mathrm{higher}\mathrm{powers}\mathrm{of}x\mathrm{are}\mathrm{negleted},$$$$\mathrm{then}$$$${(1+3x)}^{\frac{3}{x}}=e^6(1-9x)$$

Commented by mr W last updated on 28/Feb/20

$$ithinkthequestioniswrong.please$$$$checksir!$$$$f\left(x\right)={(1+3x)}^{\frac{3}{x}}$$$$g\left(x\right)=e^6(1-9x)$$$$f\left(0\right)\ne g\left(0\right)$$$$f^{\prime }\left(0\right)\ne g^{\prime }\left(0\right)$$$$\Rightarrow f\left(x\right)\approx g\left(x\right)isnottrue!$$$$$$$$ithinkcorrectis$$$${(1+3x)}^{\frac{3}{x}}\approx e^9(1-\frac{27}{2}x)$$

Commented by Rio Michael last updated on 28/Feb/20

$$thankssir,imetthat$$$$questioninanexam$$$$andiconcludedasfollows:$$$$\ln {(1+3x)}^{\frac{3}{x}}\nRightarrow {(1+3x)}^{\frac{3}{x}}\ne e^6(1-9x)$$$$hopeitslegitsir$$

Commented by mr W last updated on 28/Feb/20

$$\ln (1+3x)=\left(3x\right)-\frac{{\left(3x\right)}^2}{2}+\frac{{\left(3x\right)}^3}{3}-\frac{{\left(3x\right)}^4}{4}+...$$$$\ln (1+3x)=3x-\frac{9x^2}{2}+\frac{27x^3}{3}-\frac{81x^4}{4}+...$$$$\frac{3}{x}\ln (1+3x)=9-\frac{27x}{2}+\frac{81x^2}{3}-\frac{243x^3}{4}+...$$$${(1+3x)}^{\frac{3}{x}}=e^{\frac{3}{x}\ln (1+3x)}=e^{9-\frac{27x}{2}+\frac{81x^2}{3}-\frac{243x^3}{4}+...}$$$$\approx e^{9-\frac{27x}{2}}=e^9{e}^{-\frac{27}{2}x}=e^9(1-\frac{27}{2}x+\frac{1}{2}{\left(\frac{27}{2}x\right)}^2...)$$$$\approx e^9(1-\frac{27}{2}x)$$

Commented by Rio Michael last updated on 28/Feb/20

$$thankssir$$

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