Find the value of x which satisfies the equation: log5110x+2 +log5(x+12 ) =0

log5110x+2 +log5(x+12 ) =0 is valid when x+2>0,x+1>0 ⇒x>−1Changing base, we can write−log510x+2 +log5(x+12 ) =0,[∵logx1y=−logxy]⇒log510x+2 =log5(x+12 ) ⇒10x+2=x+12⇒x=−6,3As x>−1,∴x=3 is the solutionAns: 3

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Find the value of $x$ which satisfies the equation: $lo g_{\frac{1}{5}} \frac{x + 2}{10} + lo g_{5} (\frac{2}{x + 1}) = 0$

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Find the value of

x

which satisfies the equation:

lo g_{\frac{1}{5}} \frac{x + 2}{10} + lo g_{5} (\frac{2}{x + 1}) = 0

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Text solutionVerified

lo g_{\frac{1}{5}} \frac{x + 2}{10} + lo g_{5} (\frac{2}{x + 1}) = 0

is valid when

x + 2 > 0, x + 1 > 0

\Rightarrow x > - 1

Changing base, we can write

- lo g_{5} \frac{x + 2}{10} + lo g_{5} (\frac{2}{x + 1}) = 0, [∵ lo g_{\frac{1}{x}} y = - lo g_{x} y]

\Rightarrow lo g_{5} \frac{x + 2}{10} = lo g_{5} (\frac{2}{x + 1})

\Rightarrow \frac{x + 2}{10} = \frac{2}{x + 1} \Rightarrow x = - 6, 3

x > - 1, ∴ x = 3

is the solution

Ans: 3

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Question Text	Find the value of $x$ which satisfies the equation: $lo g_{\frac{1}{5}} \frac{x + 2}{10} + lo g_{5} (\frac{2}{x + 1}) = 0$
Answer Type	Text solution:1
Upvotes	150

Find the value of x which satisfies the equation: log51​​10x+2​ +log5​(x+12​ ) =0

Text solutionVerified

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Find the value of $x$ which satisfies the equation: $lo g_{\frac{1}{5}} \frac{x + 2}{10} + lo g_{5} (\frac{2}{x + 1}) = 0$