Make-r-the-subject-of-the-formular-S-a-r-n-1-r-1-

Question Number 62942 by Tawa1 last updated on 27/Jun/19

$$\mathrm{Make}\:\:\mathrm{r}\:\:\mathrm{the}\:\mathrm{subject}\:\mathrm{of}\:\mathrm{the}\:\mathrm{formular}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{S}\:\:=\:\:\frac{\mathrm{a}\left(\mathrm{r}^{\mathrm{n}} \:−\:\mathrm{1}\right)}{\mathrm{r}\:−\:\mathrm{1}} \\ $$

Commented by Tawa1 last updated on 27/Jun/19

Sir, help me solve this Find the value of x: 5^x + 6x = 7

$$\mathrm{Sir},\:\:\mathrm{help}\:\mathrm{me}\:\mathrm{solve}\:\mathrm{this} \\ $$$$\:\:\:\:\:\:\: \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\:\mathrm{x}:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}^{\mathrm{x}} \:+\:\mathrm{6x}\:\:=\:\:\mathrm{7} \\ $$

Commented by Tawa1 last updated on 27/Jun/19

Sir mrW can we use Lanbert function ? Or anybody help.

$$\mathrm{Sir}\:\mathrm{mrW}\:\mathrm{can}\:\mathrm{we}\:\mathrm{use}\:\mathrm{Lanbert}\:\mathrm{function}\:? \\ $$$$\mathrm{Or}\:\mathrm{anybody}\:\mathrm{help}. \\ $$

Commented by mr W last updated on 27/Jun/19

$${no} \\ $$

Commented by Tawa1 last updated on 27/Jun/19

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$

Commented by mr W last updated on 27/Jun/19

5^x +6x=7 5^x =6((7/6)−x) ((7/6)−x)5^(−x) =(1/6) ((7/6)−x)5^((7/6)−x) =(1/6)×5^(7/6) ((7/6)−x)e^(((7/6)−x)ln 5) =(1/6)×5^(7/6) [((7/6)−x)ln 5]e^(((7/6)−x)ln 5) =(1/6)×5^(7/6) ×ln 5 ⇒((7/6)−x)ln 5=W((1/6)×5^(7/6) ×ln 5) ⇒x=(7/6)−(1/(ln 5))W((1/6)×5^(7/6) ×ln 5) ⇒≈(7/6)−((0.7933)/(ln 5))=0.6737

$$\mathrm{5}^{{x}} +\mathrm{6}{x}=\mathrm{7} \\ $$$$\mathrm{5}^{{x}} =\mathrm{6}\left(\frac{\mathrm{7}}{\mathrm{6}}−{x}\right) \\ $$$$\left(\frac{\mathrm{7}}{\mathrm{6}}−{x}\right)\mathrm{5}^{−{x}} =\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\left(\frac{\mathrm{7}}{\mathrm{6}}−{x}\right)\mathrm{5}^{\frac{\mathrm{7}}{\mathrm{6}}−{x}} =\frac{\mathrm{1}}{\mathrm{6}}×\mathrm{5}^{\frac{\mathrm{7}}{\mathrm{6}}} \\ $$$$\left(\frac{\mathrm{7}}{\mathrm{6}}−{x}\right){e}^{\left(\frac{\mathrm{7}}{\mathrm{6}}−{x}\right)\mathrm{ln}\:\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{6}}×\mathrm{5}^{\frac{\mathrm{7}}{\mathrm{6}}} \\ $$$$\left[\left(\frac{\mathrm{7}}{\mathrm{6}}−{x}\right)\mathrm{ln}\:\mathrm{5}\right]{e}^{\left(\frac{\mathrm{7}}{\mathrm{6}}−{x}\right)\mathrm{ln}\:\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{6}}×\mathrm{5}^{\frac{\mathrm{7}}{\mathrm{6}}} ×\mathrm{ln}\:\mathrm{5} \\ $$$$\Rightarrow\left(\frac{\mathrm{7}}{\mathrm{6}}−{x}\right)\mathrm{ln}\:\mathrm{5}={W}\left(\frac{\mathrm{1}}{\mathrm{6}}×\mathrm{5}^{\frac{\mathrm{7}}{\mathrm{6}}} ×\mathrm{ln}\:\mathrm{5}\right) \\ $$$$\Rightarrow{x}=\frac{\mathrm{7}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{5}}\mathbb{W}\left(\frac{\mathrm{1}}{\mathrm{6}}×\mathrm{5}^{\frac{\mathrm{7}}{\mathrm{6}}} ×\mathrm{ln}\:\mathrm{5}\right) \\ $$$$\Rightarrow\approx\frac{\mathrm{7}}{\mathrm{6}}−\frac{\mathrm{0}.\mathrm{7933}}{\mathrm{ln}\:\mathrm{5}}=\mathrm{0}.\mathrm{6737} \\ $$

Commented by kaivan.ahmadi last updated on 27/Jun/19

5^x =7−6x if we plote y_1 =5^x and y_2 =7−6x then we see that they cut off each othere at one point,this means that this equation has one root. now if f(x)=5^x +6x−7=0 {: ((f(0)<0)),((f(1)>0)) }⇒f(0)f(1)<0 f((1/2))<0 so the root is between ((1/2),1) f((3/4))>0 so the root is between ((1/2),(3/4))

$$\mathrm{5}^{{x}} =\mathrm{7}−\mathrm{6}{x} \\ $$$${if}\:{we}\:{plote}\:{y}_{\mathrm{1}} =\mathrm{5}^{{x}} \:{and}\:{y}_{\mathrm{2}} =\mathrm{7}−\mathrm{6}{x}\:{then}\:{we}\:{see}\:{that} \\ $$$${they}\:{cut}\:{off}\:{each}\:{othere}\:{at}\:{one}\:{point},{this} \\ $$$${means}\:{that}\:{this}\:{equation}\:{has}\:{one}\:{root}. \\ $$$${now}\:{if} \\ $$$${f}\left({x}\right)=\mathrm{5}^{{x}} +\mathrm{6}{x}−\mathrm{7}=\mathrm{0} \\ $$$$ \\ $$$$\left.\begin{matrix}{{f}\left(\mathrm{0}\right)<\mathrm{0}}\\{{f}\left(\mathrm{1}\right)>\mathrm{0}}\end{matrix}\right\}\Rightarrow{f}\left(\mathrm{0}\right){f}\left(\mathrm{1}\right)<\mathrm{0} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)<\mathrm{0} \\ $$$${so}\:{the}\:{root}\:{is}\:{between}\:\left(\frac{\mathrm{1}}{\mathrm{2}},\mathrm{1}\right) \\ $$$${f}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)>\mathrm{0} \\ $$$${so}\:{the}\:{root}\:{is}\:{between}\:\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Hope last updated on 27/Jun/19

f(x)=5^x +6x−7 f(0)<0 f(1)=5^1 +6−7>0 f(0.5)=5^(0.5) +6×0.5−7 =2.23+3−7<0 f(0.6)=5^(0.6) +6×0.6−7 =2.63+3.6−7<0 f(0.7)=5^(0.7) +6×0.7−7 =3.09+4.2−7>0 so value of x 0.7>x>0.6

$${f}\left({x}\right)=\mathrm{5}^{{x}} +\mathrm{6}{x}−\mathrm{7} \\ $$$${f}\left(\mathrm{0}\right)<\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{5}^{\mathrm{1}} +\mathrm{6}−\mathrm{7}>\mathrm{0} \\ $$$${f}\left(\mathrm{0}.\mathrm{5}\right)=\mathrm{5}^{\mathrm{0}.\mathrm{5}} +\mathrm{6}×\mathrm{0}.\mathrm{5}−\mathrm{7} \\ $$$$\:\:\:\:=\mathrm{2}.\mathrm{23}+\mathrm{3}−\mathrm{7}<\mathrm{0} \\ $$$${f}\left(\mathrm{0}.\mathrm{6}\right)=\mathrm{5}^{\mathrm{0}.\mathrm{6}} +\mathrm{6}×\mathrm{0}.\mathrm{6}−\mathrm{7} \\ $$$$=\mathrm{2}.\mathrm{63}+\mathrm{3}.\mathrm{6}−\mathrm{7}<\mathrm{0} \\ $$$${f}\left(\mathrm{0}.\mathrm{7}\right)=\mathrm{5}^{\mathrm{0}.\mathrm{7}} +\mathrm{6}×\mathrm{0}.\mathrm{7}−\mathrm{7} \\ $$$$\:\:\:=\mathrm{3}.\mathrm{09}+\mathrm{4}.\mathrm{2}−\mathrm{7}>\mathrm{0} \\ $$$${so}\:{value}\:{of}\:{x}\:\:\:\:\mathrm{0}.\mathrm{7}>{x}>\mathrm{0}.\mathrm{6} \\ $$

Commented by Tawa1 last updated on 27/Jun/19