Question Number 35491 by mondodotto@gmail.com last updated on 19/May/18
$$\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}} \\ $$$$\mathrm{3}^{\boldsymbol{{x}}} =\mathrm{9}\boldsymbol{{x}} \\ $$
Commented by math khazana by abdo last updated on 12/Jun/18
$${the}\:{question}\:{is}\:{not}\:{correct}\:{its}\:{only}\:{a}\:{equation}\: \\ $$$${because}\:{y}=\mathrm{9}{x}\:{is}\:{equation}\:{of}\:{a}\:{line}\:\:{y}=\mathrm{3}^{{x}} \:={e}^{{xln}\left(\mathrm{3}\right)} \\ $$$${never}\:{be}\:{a}\:{equation}\:{of}\:{a}\:{line}\:! \\ $$
Answered by candre last updated on 19/May/18
$${x}=\mathrm{0}\Rightarrow\mathrm{3}^{{x}} =\mathrm{1}\neq\mathrm{0}=\mathrm{9}{x} \\ $$$$\mathrm{therefore}\:\mathrm{afirmation}\:\mathrm{is}\:\mathrm{false} \\ $$
Commented by Rasheed.Sindhi last updated on 19/May/18
$$\mathrm{This}\:\mathrm{is}\:\mathrm{a}\:\mathrm{conditional}\:\mathrm{equation}.\mathrm{Therfore} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{satisfied}\:\mathrm{by}\:\mathrm{specific}\:\mathrm{value}/\mathrm{s}. \\ $$
Commented by candre last updated on 20/May/18
$$\mathrm{but}\:\mathrm{if}\:\mathrm{it}\:\mathrm{was}\:\mathrm{that}\:\mathrm{would}\:\mathrm{it}\:\mathrm{be} \\ $$$$“\mathrm{find}\:{x}\:\mathrm{such}\:\mathrm{afirmation}\:\mathrm{is}\:\mathrm{true}''? \\ $$
Commented by Rasheed.Sindhi last updated on 20/May/18
$$\mathrm{Sir}\:\mathrm{you}\:\mathrm{are}\:\mathrm{right},\mathrm{the}\:\mathrm{questioner} \\ $$$$\mathrm{has}\:\mathrm{written}\:“\mathrm{Prove}\:\mathrm{that}''.\mathrm{I}\:\mathrm{didn}'\mathrm{t} \\ $$$$\mathrm{consider}\:\mathrm{it}\:\mathrm{at}\:\mathrm{first}. \\ $$
Answered by rahul 19 last updated on 20/May/18
$$\mathrm{3}^{\mathrm{3}} =\:\mathrm{9}×\mathrm{3} \\ $$$$\Rightarrow\:{x}=\mathrm{3}. \\ $$
Answered by MrW3 last updated on 12/Jun/18
$$\mathrm{3}^{{x}} =\mathrm{9}{x} \\ $$$${e}^{{x}\mathrm{ln}\:\mathrm{3}} =\mathrm{9}{x} \\ $$$$\mathrm{9}{xe}^{−{x}\mathrm{ln}\:\mathrm{3}} =\mathrm{1} \\ $$$$\left(−{x}\mathrm{ln}\:\mathrm{3}\right){e}^{−{x}\mathrm{ln}\:\mathrm{3}} =−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{9}} \\ $$$$\Rightarrow−{x}\mathrm{ln}\:\mathrm{3}={W}\left(−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{9}}\right) \\ $$$$\Rightarrow{x}=−\frac{{W}\left(−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{9}}\right)}{\mathrm{ln}\:\mathrm{3}}=\begin{cases}{\mathrm{0}.\mathrm{1279}}\\{\mathrm{3}}\end{cases} \\ $$