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Question Number 192458 by Mingma last updated on 18/May/23

	Answered by Frix last updated on 19/May/23
	$log_(x+(7/2)) (((x+7)/(2x+3)))^2 =((ln (((x+7)/(2x+3)))^2 )/(ln (x+(7/2))))= =((2(ln ∣x+7∣ −ln ∣2x+3∣))/(ln (2x+7) −ln 2))≥0 ((ln ∣x+7∣ −ln ∣2x+3∣)/(ln (2x+7) −ln 2))≥0 2x+7>0 ⇔ x>−(7/2) ln (2x+7) −ln 2 ≠0 ⇔ x≠−(5/2) [1] ln (2x+7) −ln 2 <0 −(7/2)<x<−(5/2); x∈Z ⇒ x=−3 Testing ln ∣x+7∣ −ln ∣2x+3∣≤0 ln 4 −ln 3 ≤0 wrong [2] ln (2x+7) −ln 2 >0 x>−(5/2) ln ∣x+7∣ −ln ∣2x+3∣≥0 ln ∣x+7∣ ≥ln ∣2x+3∣ x≠−7∧x≠−(3/2) ⇒ −((10)/3)≤x≤4 x∈Z ⇒ x∈{−3, −2, −1, 0, 1, 2, 3, 4} 8 integer solutions$
	$\log_{x + \frac{7}{2}} {(\frac{x + 7}{2 x + 3})}^{2} = \frac{\ln {(\frac{x + 7}{2 x + 3})}^{2}}{\ln (x + \frac{7}{2})} =$ $= \frac{2 (\ln ∣ x + 7 ∣ - \ln ∣ 2 x + 3 ∣)}{\ln (2 x + 7) - \ln 2} ⩾ 0$ $\frac{\ln ∣ x + 7 ∣ - \ln ∣ 2 x + 3 ∣}{\ln (2 x + 7) - \ln 2} ⩾ 0$ $2 x + 7 > 0 \Leftrightarrow x > - \frac{7}{2}$ $\ln (2 x + 7) - \ln 2 \neq 0 \Leftrightarrow x \neq - \frac{5}{2}$ $[1]$ $\ln (2 x + 7) - \ln 2 < 0$ $- \frac{7}{2} < x < - \frac{5}{2}; x \in Z \Rightarrow x = - 3$ $Testing$ $\ln ∣ x + 7 ∣ - \ln ∣ 2 x + 3 ∣⩽ 0$ $\ln 4 - \ln 3 ⩽ 0 wrong$ $[2]$ $\ln (2 x + 7) - \ln 2 > 0$ $x > - \frac{5}{2}$ $\ln ∣ x + 7 ∣ - \ln ∣ 2 x + 3 ∣⩾ 0$ $\ln ∣ x + 7 ∣ ⩾ \ln ∣ 2 x + 3 ∣$ $x \neq - 7 \land x \neq - \frac{3}{2}$ $\Rightarrow$ $- \frac{10}{3} ⩽ x ⩽ 4$ $x \in Z \Rightarrow x \in {- 3, - 2, - 1, 0, 1, 2, 3, 4}$ $8 integer solutions$
	Commented by Mingma last updated on 19/May/23
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