Menu Close

y-x-64-y-x-1-x-1-16-find-x-




Question Number 184019 by HeferH last updated on 02/Jan/23
 { ((y^x = 64)),((y^((x + 1)/(x − 1))  = 16)) :}   find “x”
$$\begin{cases}{{y}^{{x}} =\:\mathrm{64}}\\{{y}^{\frac{{x}\:+\:\mathrm{1}}{{x}\:−\:\mathrm{1}}} \:=\:\mathrm{16}}\end{cases} \\ $$$$\:{find}\:“{x}'' \\ $$
Commented by a.lgnaoui last updated on 02/Jan/23
Hapy New year
$${Hapy}\:{New}\:{year} \\ $$
Commented by Gazella thomsonii last updated on 02/Jan/23
=Ha^2 e^2 p^2 wy^2 r
$$=\mathrm{Ha}^{\mathrm{2}} \mathrm{e}^{\mathrm{2}} \mathrm{p}^{\mathrm{2}} \mathrm{wy}^{\mathrm{2}} \mathrm{r} \\ $$
Answered by manolex last updated on 02/Jan/23
xlny=3ln4  ((x+1)/(x−1))lny=2ln4  x=((3(x+1))/(2(x−1)))  2x^2 −2x=3x+3  2x^2 −5x−3=0  2x               +1  x                  −3  2x+1=0     x=−(1/2)  x−3=0       x=3  comprobando  x_1 =3     x_2 =−(1/2)
$${xlny}=\mathrm{3}{ln}\mathrm{4} \\ $$$$\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}{lny}=\mathrm{2}{ln}\mathrm{4} \\ $$$${x}=\frac{\mathrm{3}\left({x}+\mathrm{1}\right)}{\mathrm{2}\left({x}−\mathrm{1}\right)} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}=\mathrm{3}{x}+\mathrm{3} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{2}{x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{1} \\ $$$${x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{3} \\ $$$$\mathrm{2}{x}+\mathrm{1}=\mathrm{0}\:\:\:\:\:{x}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}−\mathrm{3}=\mathrm{0}\:\:\:\:\:\:\:{x}=\mathrm{3} \\ $$$${comprobando} \\ $$$${x}_{\mathrm{1}} =\mathrm{3}\:\:\:\:\:{x}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by manolex last updated on 02/Jan/23
Happy new year,Sr.HeferH
$${Happy}\:{new}\:{year},{Sr}.{HeferH} \\ $$
Commented by HeferH last updated on 02/Jan/23
Happy new year !   :)
$${Happy}\:{new}\:{year}\:! \\ $$$$\left.\::\right) \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *