Solution set of log 0-25 x 2-2 x -82-log 0-510-3 x - x 2-1A- 1-6-313-1- 1-2-73-7B- -3C- 1-4D- 1-7

The correct option is A {1/6(√(313) 1),1/2(√(73) 7)} log_0.25(x^2+2x 8)^2 log_0.5(10+3x+x^2)=1 can be rewriten as log_0.5|x^2+2x 8|/10+3x x^2=1 ⇒|x^2+2x 8|/10 ...

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Standard IX

Mathematics

Inequalities Involving Modulus Function

Solution set ...

Question

Solution set of $l o g_{0.25} (x^{2} + 2 x - 8)^{2} - l o g_{0.5} (10 + 3 x + x^{2}) = 1$

${- 3}$

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${\frac{1}{4}}$

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${\frac{1}{6} (\sqrt{313} - 1), \frac{1}{2} (\sqrt{73} - 7)}$

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${1, 7}$

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Solution

The correct option is C
${\frac{1}{6} (\sqrt{313} - 1), \frac{1}{2} (\sqrt{73} - 7)}$

$l o g_{0.25} (x^{2} + 2 x - 8)^{2} - l o g_{0.5} (10 + 3 x + x^{2}) = 1$ can be rewriten as
$l o g_{0.5} \frac{| x^{2} + 2 x - 8 |}{10 + 3 x - x^{2}} = 1 \Rightarrow \frac{| x^{2} + 2 x - 8 |}{10 + 3 x - x^{2}} = \frac{1}{2} \Rightarrow 2 | x^{2} + 2 x - 8 | = 10 + 3 x - x^{2} \Rightarrow 2 | (x + 4) (x - 2) | = (5 - x) (2 + x) \Rightarrow x = \frac{1}{6} (\sqrt{313} - 1), \frac{1}{2} (\sqrt{73} - 7)$

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Similar questions

Solution set of $l o g_{0.25} (x^{2} + 2 x - 8)^{2} - l o g_{0.5} (10 + 3 x + x^{2}) = 1$

Q. Match the equations in List 1 with the solutions in List 2

	List I		List II
A.	${log}_{0.25} {(x^{2} + 2 x - 8)}^{2}$ $- {log}_{0.5} (10 + 3 x - x^{2}) = 1$	1.	${- 3}$
B.	${log}_{2} (x^{2} + 7)$ $= 5 + {log}_{2} x$ $- \frac{6}{{log}_{2} (x + \frac{7}{x})}$	2.	${\frac{1}{4}}$
C.	${log}_{(1 - 2 x)} (6 x^{2} - 5 x + 1)$ $- {log}_{(1 - 3 x)} (4 x^{2} - 4 x + 1) = 2$	3.	${\frac{1}{6} (\sqrt{313} - 1), \frac{1}{2} (\sqrt{73} - 7)}$
D.	${log}_{10} (1 {+ x}^{2} - 2 x) + 1 - {log}_{10} (1 + x^{2}) =$ $2 {log}_{10} (1 - x)$	4.	${1, 7}$

The set $A = {x : x ϵ R, x^{2} = 16 a n d 2 x = 6}$ equals

Q. Find the square root of

\sqrt{2 x} + \sqrt{- x^{4} - 1}

Q. The number of solutions of the equation

1 + {sin}^{4} x = {cos}^{2} 3 x, x \in [- \frac{5 π}{2}, \frac{5 π}{2}]

is :

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Solution set of log0.25(x2+2x−8)2−log0.5(10+3x+x2)=1

The correct option is C {16(√313−1),12(√73−7)} log0.25(x2+2x−8)2−log0.5(10+3x+x2)=1 can be rewriten as log0.5|x2+2x−8|10+3x−x2=1⇒|x2+2x−8|10+3x−x2=12⇒2|x2+2x−8|=10+3x−x2⇒2|(x+4)(x−2)|=(5−x)(2+x)⇒x=16(√313−1),12(√73−7)

Solution set of $l o g_{0.25} (x^{2} + 2 x - 8)^{2} - l o g_{0.5} (10 + 3 x + x^{2}) = 1$