Question Number 1343 by Rasheed Soomro last updated on 24/Jul/15
$$\mathrm{3}^{{log}\:\mathrm{3}{x}+\mathrm{4}} =\mathrm{4}^{{log}\:\mathrm{4}{x}+\mathrm{3}} \\ $$
Answered by 112358 last updated on 24/Jul/15
![Taking logs to base e on both sides ⇒(log3x+4)ln3=(log4x+3)ln4 (logx+log3+4)ln3=(logx+log4+3)ln4 let p=logx ∴ pln3+(log3+4)ln3=pln4+(log4+3)ln4 p(ln3−ln4)=(log4+3)ln4−(log3+4)ln3 p=(((log4+3)ln4−(log3+4)ln3)/(ln(3/4))) Since p=logx ⇒ logx=(((log4+3)ln4−(log3+4)ln3)/(ln(3/4))) ∴ x=10^(((log4+3)ln4−(log3+4)ln3)/(ln(3/4))) x≈0.549111367 Alternatively Plotting the graph of y=3^(log3x+4) −4^(log4x+3) and reading off its real root gives an approximate solution for x satisfying the given equation. If reading off does not yield an accurate result other numerical methods like Newton− Raphson′s iterative formular, linear interpolation or interval bisection can give better estimations if we define ∃x[0,1]∣y=0 .](https://www.tinkutara.com/question/Q1344.png)
$${Taking}\:{logs}\:{to}\:{base}\:{e}\:{on}\:{both}\:{sides} \\ $$$$\Rightarrow\left({log}\mathrm{3}{x}+\mathrm{4}\right){ln}\mathrm{3}=\left({log}\mathrm{4}{x}+\mathrm{3}\right){ln}\mathrm{4} \\ $$$$\left({logx}+{log}\mathrm{3}+\mathrm{4}\right){ln}\mathrm{3}=\left({logx}+{log}\mathrm{4}+\mathrm{3}\right){ln}\mathrm{4} \\ $$$${let}\:{p}={logx}\: \\ $$$$\therefore\:{pln}\mathrm{3}+\left({log}\mathrm{3}+\mathrm{4}\right){ln}\mathrm{3}={pln}\mathrm{4}+\left({log}\mathrm{4}+\mathrm{3}\right){ln}\mathrm{4} \\ $$$${p}\left({ln}\mathrm{3}−{ln}\mathrm{4}\right)=\left({log}\mathrm{4}+\mathrm{3}\right){ln}\mathrm{4}−\left({log}\mathrm{3}+\mathrm{4}\right){ln}\mathrm{3} \\ $$$${p}=\frac{\left({log}\mathrm{4}+\mathrm{3}\right){ln}\mathrm{4}−\left({log}\mathrm{3}+\mathrm{4}\right){ln}\mathrm{3}}{{ln}\left(\mathrm{3}/\mathrm{4}\right)} \\ $$$${Since}\:{p}={logx} \\ $$$$\Rightarrow\:{logx}=\frac{\left({log}\mathrm{4}+\mathrm{3}\right){ln}\mathrm{4}−\left({log}\mathrm{3}+\mathrm{4}\right){ln}\mathrm{3}}{{ln}\left(\mathrm{3}/\mathrm{4}\right)} \\ $$$$\therefore\:{x}=\mathrm{10}^{\frac{\left({log}\mathrm{4}+\mathrm{3}\right){ln}\mathrm{4}−\left({log}\mathrm{3}+\mathrm{4}\right){ln}\mathrm{3}}{{ln}\left(\mathrm{3}/\mathrm{4}\right)}} \\ $$$${x}\approx\mathrm{0}.\mathrm{549111367} \\ $$$${Alternatively} \\ $$$${Plotting}\:{the}\:{graph}\:{of}\: \\ $$$${y}=\mathrm{3}^{{log}\mathrm{3}{x}+\mathrm{4}} −\mathrm{4}^{{log}\mathrm{4}{x}+\mathrm{3}} \:{and}\:{reading}\:{off} \\ $$$${its}\:{real}\:{root}\:{gives}\:{an}\:{approximate} \\ $$$${solution}\:{for}\:{x}\:{satisfying}\:{the}\:{given} \\ $$$${equation}.\:{If}\:{reading}\:{off}\:{does}\:{not} \\ $$$${yield}\:{an}\:{accurate}\:{result}\:{other}\: \\ $$$${numerical}\:{methods}\:{like}\:{Newton}− \\ $$$${Raphson}'{s}\:{iterative}\:{formular}, \\ $$$${linear}\:{interpolation}\:{or}\:{interval} \\ $$$${bisection}\:{can}\:{give}\:{better}\:{estimations} \\ $$$${if}\:{we}\:{define}\:\exists{x}\left[\mathrm{0},\mathrm{1}\right]\mid{y}=\mathrm{0}\:. \\ $$
Commented by Rasheed Ahmad last updated on 24/Jul/15
$${Good}\:{work}! \\ $$