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Q1-a-solve-for-x-9-x-5-3-x-6-b-write-down-the-first-4-terms-in-the-binomial-expansion-of-1-3x-7-c-the-sum-S-n-of-the-first-n-th-terms-is-given-by-S-n-3-1-2-3-n-find-d-the-c

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Question Number 35304 by Rio Mike last updated on 17/May/18
Q1. a) solve for x 9^x +5(3^x )=6 b)write down the first 4 terms in the binomial expansion of (1−3x)^7 c)the sum S_n of the first n^(th) terms is given by S_(n ) = 3(1−((2/3))^n ) find d) the common ratio e) the sum to infinity of the progression
$$\left.\:{Q}\mathrm{1}.\:\:\:{a}\right)\:{solve}\:{for}\:{x}\:\:\mathrm{9}^{{x}} +\mathrm{5}\left(\mathrm{3}^{{x}} \right)=\mathrm{6} \\ $$$$\left.{b}\right){write}\:{down}\:{the}\:{first}\:\:\mathrm{4}\:{terms} \\ $$$${in}\:{the}\:{binomial}\:{expansion}\:{of}\:\left(\mathrm{1}−\mathrm{3}{x}\right)^{\mathrm{7}} \\ $$$$\left.{c}\right){the}\:{sum}\:{S}_{{n}} \:{of}\:{the}\:{first}\:{n}^{{th}} {terms} \\ $$$${is}\:{given}\:{by}\:{S}_{{n}\:} =\:\mathrm{3}\left(\mathrm{1}−\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{n}} \right)\:{find} \\ $$$$\left.{d}\right)\:{the}\:{common}\:{ratio} \\ $$$$\left.{e}\right)\:{the}\:{sum}\:{to}\:{infinity}\:{of}\:{the}\:{progression} \\ $$
Commented by prakash jain last updated on 17/May/18
9^x +5∙3^x =6 3^(2x) +5∙3^x −6=0 3^(2x) +6∙3^x −3^x −6=0 3^x (3^x +6)−(3^x +6)=0 (3^x −1)(3^x +6)=0 3^x =−6 (not possible fof x∈R 3^x =1⇒x=0 ans x=0
$$\mathrm{9}^{{x}} +\mathrm{5}\centerdot\mathrm{3}^{{x}} =\mathrm{6} \\ $$$$\mathrm{3}^{\mathrm{2}{x}} +\mathrm{5}\centerdot\mathrm{3}^{{x}} −\mathrm{6}=\mathrm{0} \\ $$$$\mathrm{3}^{\mathrm{2}{x}} +\mathrm{6}\centerdot\mathrm{3}^{{x}} −\mathrm{3}^{{x}} −\mathrm{6}=\mathrm{0} \\ $$$$\mathrm{3}^{{x}} \left(\mathrm{3}^{{x}} +\mathrm{6}\right)−\left(\mathrm{3}^{{x}} +\mathrm{6}\right)=\mathrm{0} \\ $$$$\left(\mathrm{3}^{{x}} −\mathrm{1}\right)\left(\mathrm{3}^{{x}} +\mathrm{6}\right)=\mathrm{0} \\ $$$$\mathrm{3}^{{x}} =−\mathrm{6}\:\left({not}\:{possible}\:{fof}\:{x}\in\mathbb{R}\right. \\ $$$$\mathrm{3}^{{x}} =\mathrm{1}\Rightarrow{x}=\mathrm{0} \\ $$$${ans}\:{x}=\mathrm{0} \\ $$
Commented by Rasheed.Sindhi last updated on 18/May/18
Welcome back SIR! I haven′t seen your post for long time.Are you too busy nowadays?
$$\mathrm{Welcome}\:\mathrm{back}\:\mathrm{SIR}!\:\mathrm{I}\:\mathrm{haven}'\mathrm{t}\:\mathrm{seen}\:\mathrm{your} \\ $$$$\mathrm{post}\:\mathrm{for}\:\mathrm{long}\:\mathrm{time}.\mathrm{Are}\:\mathrm{you}\:\mathrm{too}\:\mathrm{busy}\: \\ $$$$\mathrm{nowadays}? \\ $$
Commented by rahul 19 last updated on 18/May/18
No problem sir ��
Commented by prakash jain last updated on 18/May/18
Was very busy in office. Could only read the post once in a week.
$$\mathrm{Was}\:\mathrm{very}\:\mathrm{busy}\:\mathrm{in}\:\mathrm{office}.\:\mathrm{Could} \\ $$$$\mathrm{only}\:\mathrm{read}\:\mathrm{the}\:\mathrm{post}\:\mathrm{once}\:\mathrm{in}\:\mathrm{a}\:\mathrm{week}. \\ $$
Answered by MJS last updated on 18/May/18
(3^x )^2 +5(3^x )−6=0 (3^x )=−(5/2)±(√(((25)/4)+6))=−(5/2)±(7/2) (3^x )=−6 ∨ (3^x )=1 x_2 =0 x_1 =((ln 6)/(ln 3))+(((2n+1)π)/(ln 3))i; n∈N_0 (a−b)^7 = ((7),(0) )×a^7 − ((7),(1) )×a^6 b+ ((7),(2) )×a^5 b^2 − ((7),(3) )×a^4 b^3 + ((7),(4) )×a^3 b^4 − ((7),(5) )×a^2 b^5 + ((7),(6) )×ab^6 − ((7),(7) )×ib^7 (1−3x)^7 =1−21x+189x^2 −945x^3 +2835x^4 −5103x^5 +5103x^6 −2187x^7 the ratio is (2/3) p_n =((2/3))^n p_n =q^n ⇒ S_n =((q^(n+1) −1)/(q−1)) S_∞ =(1/(1−q))=3
$$\left(\mathrm{3}^{{x}} \right)^{\mathrm{2}} +\mathrm{5}\left(\mathrm{3}^{{x}} \right)−\mathrm{6}=\mathrm{0} \\ $$$$\left(\mathrm{3}^{{x}} \right)=−\frac{\mathrm{5}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{25}}{\mathrm{4}}+\mathrm{6}}=−\frac{\mathrm{5}}{\mathrm{2}}\pm\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\left(\mathrm{3}^{{x}} \right)=−\mathrm{6}\:\vee\:\left(\mathrm{3}^{{x}} \right)=\mathrm{1} \\ $$$${x}_{\mathrm{2}} =\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{ln}\:\mathrm{6}}{\mathrm{ln}\:\mathrm{3}}+\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{ln}\:\mathrm{3}}\mathrm{i};\:{n}\in\mathbb{N}_{\mathrm{0}} \\ $$$$ \\ $$$$\left({a}−{b}\right)^{\mathrm{7}} =\begin{pmatrix}{\mathrm{7}}\\{\mathrm{0}}\end{pmatrix}×{a}^{\mathrm{7}} −\begin{pmatrix}{\mathrm{7}}\\{\mathrm{1}}\end{pmatrix}×{a}^{\mathrm{6}} {b}+\begin{pmatrix}{\mathrm{7}}\\{\mathrm{2}}\end{pmatrix}×{a}^{\mathrm{5}} {b}^{\mathrm{2}} −\begin{pmatrix}{\mathrm{7}}\\{\mathrm{3}}\end{pmatrix}×{a}^{\mathrm{4}} {b}^{\mathrm{3}} +\begin{pmatrix}{\mathrm{7}}\\{\mathrm{4}}\end{pmatrix}×{a}^{\mathrm{3}} {b}^{\mathrm{4}} −\begin{pmatrix}{\mathrm{7}}\\{\mathrm{5}}\end{pmatrix}×{a}^{\mathrm{2}} {b}^{\mathrm{5}} +\begin{pmatrix}{\mathrm{7}}\\{\mathrm{6}}\end{pmatrix}×{ab}^{\mathrm{6}} −\begin{pmatrix}{\mathrm{7}}\\{\mathrm{7}}\end{pmatrix}×{ib}^{\mathrm{7}} \\ $$$$\left(\mathrm{1}−\mathrm{3}{x}\right)^{\mathrm{7}} =\mathrm{1}−\mathrm{21}{x}+\mathrm{189}{x}^{\mathrm{2}} −\mathrm{945}{x}^{\mathrm{3}} +\mathrm{2835}{x}^{\mathrm{4}} −\mathrm{5103}{x}^{\mathrm{5}} +\mathrm{5103}{x}^{\mathrm{6}} −\mathrm{2187}{x}^{\mathrm{7}} \\ $$$$ \\ $$$$\mathrm{the}\:\mathrm{ratio}\:\mathrm{is}\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${p}_{{n}} =\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{n}} \\ $$$${p}_{{n}} ={q}^{{n}} \:\Rightarrow\:{S}_{{n}} =\frac{{q}^{{n}+\mathrm{1}} −\mathrm{1}}{{q}−\mathrm{1}} \\ $$$${S}_{\infty} =\frac{\mathrm{1}}{\mathrm{1}−{q}}=\mathrm{3} \\ $$
Answered by Rasheed.Sindhi last updated on 27/May/18
(c) S_n = 3(1−((2/3))^n ) First term=a=3(1−((2/3))^1 )=1 ⋇Common ratio r 1+r=(S_2 /S_1 )=((3(1−((2/3))^2 ))/(3(1−((2/3))^1 ))) [(S_2 /S_1 )=((a+ar)/a)=1+r] =(((1−(2/3))(1+(2/3)))/((1−(2/3))))=(5/3) r=(5/3)−1=(2/3) ⋇ Sum of infinite terms=(a/(1−r))=(1/(1−(2/3))) =(1/(1/3))=3
$$\left(\mathrm{c}\right)\:{S}_{\mathrm{n}} =\:\mathrm{3}\left(\mathrm{1}−\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{n}} \right) \\ $$$$\:\:\mathrm{First}\:\mathrm{term}={a}=\mathrm{3}\left(\mathrm{1}−\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{1}} \right)=\mathrm{1} \\ $$$$\:\:\:\divideontimes\mathrm{Common}\:\mathrm{ratio}\:\:{r} \\ $$$$\mathrm{1}+{r}=\frac{{S}_{\mathrm{2}} }{{S}_{\mathrm{1}} }=\frac{\mathrm{3}\left(\mathrm{1}−\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} \right)}{\mathrm{3}\left(\mathrm{1}−\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{1}} \right)}\:\:\:\left[\frac{\mathrm{S}_{\mathrm{2}} }{\mathrm{S}_{\mathrm{1}} }=\frac{{a}+{ar}}{{a}}=\mathrm{1}+{r}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}\right)}{\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}\right)}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\mathrm{r}=\frac{\mathrm{5}}{\mathrm{3}}−\mathrm{1}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\:\:\:\:\divideontimes\:\mathrm{Sum}\:\mathrm{of}\:\mathrm{infinite}\:\mathrm{terms}=\frac{{a}}{\mathrm{1}−{r}}=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{1}/\mathrm{3}}=\mathrm{3} \\ $$
Commented by Rasheed.Sindhi last updated on 27/May/18
Corrected.
$$\mathrm{Corrected}. \\ $$

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