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Question Number 121201 by Ar Brandon last updated on 05/Nov/20

Commented by Ar Brandon last updated on 06/Nov/20

Commented by MJS_new last updated on 06/Nov/20

$I get 2 / 2 / 1 / 3 solutions$ $A r B r C t D q$
Commented by MJS_new last updated on 06/Nov/20

$B is a special case because at x = 0 both$ $logarithms are complex but the imaginary$ $part cancels out . strictly real calculation$ $leads to only one solution$
Commented by Ar Brandon last updated on 06/Nov/20
Thanks Sir. For D after solving I had 3 solutions then rejected 3–√2 making them 2 solutions. Did you notice that too ? Or I made the mistake ?
Commented by liberty last updated on 06/Nov/20

$D have double roots x_{1} = x_{2} = 3 .$ $then the number of solution is 3$
Commented by MJS_new last updated on 06/Nov/20

${it}^{'} s not clear . you could say a double root$ $is only one solution . . .$
	Answered by liberty last updated on 06/Nov/20
	$(2c) log _3 (3^x −8)=log _3 (3^(2−x) ) ⇒ numerus 3^x >8⇒x>3^(log _3 (8)) ⇔ 3^x −8=(9/3^x ) ; (3^x )^2 −8.3^x −9=0 ⇔ (3^x +1)(3^x −9)=0→ { ((3^x =9⇒x=2)),((3^x =−1(rejected))) :} The number of solution is 1$
	$(2 c) \log_{3} (3^{x} - 8) = \log_{3} (3^{2 - x})$ $\Rightarrow numerus 3^{x} > 8 \Rightarrow x > 3^{\log_{3} (8)}$ $\Leftrightarrow 3^{x} - 8 = \frac{9}{3^{x}}; {(3^{x})}^{2} - 8 . 3^{x} - 9 = 0$ $\Leftrightarrow (3^{x} + 1) (3^{x} - 9) = 0 \to {\begin{cases} 3^{x} = 9 \Rightarrow x = 2 \\ 3^{x} = - 1 (rejected) \end{cases}$ $The number of solution is 1$
	Answered by liberty last updated on 06/Nov/20
	$(2A) log _(10) (3x^2 +12x+19)−log _(10) (3x+4)=1 ⇒log _(10) (3x^2 +12x+19)=log _(10) (30x+40) ⇒3x^2 +12x+19−30x−40=0 ⇒3x^2 −18x−21=0 ⇒x^2 −6x−7=0 ;(x−7)(x+1)=0 { ((x=7)),((x=−1→ { ((3.(−1)+4>0)),((3(−1)^2 +12(−1)+19>0)) :})) :} The number of solution is 2$
	$(2 A) \log_{10} ({3 x}^{2} + 12 x + 19) - \log_{10} (3 x + 4) = 1$ $\Rightarrow \log_{10} ({3 x}^{2} + 12 x + 19) = \log_{10} (30 x + 40)$ $\Rightarrow {3 x}^{2} + 12 x + 19 - 30 x - 40 = 0$ $\Rightarrow {3 x}^{2} - 18 x - 21 = 0$ $\Rightarrow x^{2} - 6 x - 7 = 0; (x - 7) (x + 1) = 0$ ${\begin{cases} x = 7 \\ x = - 1 \to {\begin{cases} 3 . (- 1) + 4 > 0 \\ 3 {(- 1)}^{2} + 12 (- 1) + 19 > 0 \end{cases} \end{cases}$ $The number of solution is 2$
	Answered by liberty last updated on 06/Nov/20
	$(2D)2log _3 (x−2)+log _3 (x−4)^2 =0 numerus { ((x>2)),((x≠4)) :} ⇔ x∈(2,4) ∪(4,∞) ⇒log _3 (x−2)^2 +log _3 (x−4)^2 =0 ⇒log _3 [(x−2)(x−4)]^2 = log _3 (1) ⇒[(x−2)(x−4)]^2 =1 ⇒(x^2 −6x+8)^2 −1=0 ⇒(x^2 −6x+9)(x^2 −6x+7)=0 { (((x−3)^2 =0⇒x=3 ←double roots)),((x=((6±(√(36−28)))/2)=((6±2(√2))/2)=3±(√2))) :} 3+(√2) 4.414214 3−(√2) (rejected) 1.585786 The number of solution is 3$
	$(2 D) 2 \log_{3} (x - 2) + \log_{3} {(x - 4)}^{2} = 0$ $numerus {\begin{cases} x > 2 \\ x \neq 4 \end{cases} \Leftrightarrow x \in (2, 4) \cup (4, \infty)$ $\Rightarrow \log_{3} {(x - 2)}^{2} + \log_{3} {(x - 4)}^{2} = 0$ $\Rightarrow \log_{3} {[(x - 2) (x - 4)]}^{2} = \log_{3} (1)$ $\Rightarrow {[(x - 2) (x - 4)]}^{2} = 1$ $\Rightarrow {(x^{2} - 6 x + 8)}^{2} - 1 = 0$ $\Rightarrow (x^{2} - 6 x + 9) (x^{2} - 6 x + 7) = 0$ ${\begin{cases} {(x - 3)}^{2} = 0 \Rightarrow x = 3 \leftarrow double roots \\ x = \frac{6 \pm \sqrt{36 - 28}}{2} = \frac{6 \pm 2 \sqrt{2}}{2} = 3 \pm \sqrt{2} \end{cases}$ $3 + \sqrt{2}$ $4.414214$ $3 - \sqrt{2} (rejected)$ $1.585786$ $The number of solution is 3$
	Commented by Ar Brandon last updated on 06/Nov/20
	Thanks�� Please can you have a look at the answers suggested by the book ?
	Commented by Ar Brandon last updated on 06/Nov/20
	There seem to be some contradictions I think. I too got some answers similar to yours but later on I had some doubts due to these contradictions
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