$\begin{aligned} If \log_{5} (x^{2} + x) - \log_{5} (x + 1) = 2, \\ then the value of x is - \end{aligned}$ $\eqalign{ & {\text{If }}{\log _5}\left( {{x^2} + x} \right) - {\log _5}\left( {x + 1} \right) = 2, \cr & {\text{then the value of }}x{\text{ is - }} \cr}$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \log_{5} (x^{2} + x) - \log_{5} (x + 1) = 2 \\ \Rightarrow \log_{5} (\frac{x^{2} + x}{x + 1}) = 2 \\ \Rightarrow \log_{5} [\frac{x (x + 1)}{x + 1}] = 2 \\ \Rightarrow \log_{5} x = 2 \\ \Rightarrow x = 5^{2} \\ = 25 \end{aligned}

$\eqalign{ & {\log _5}\left( {{x^2} + x} \right) - {\log _5}\left( {x + 1} \right) = 2 \cr & \Rightarrow {\log _5}\left( {\frac{{{x^2} + x}}{{x + 1}}} \right) = 2\, \cr & \Rightarrow {\log _5}\left[ {\frac{{x\left( {x + 1} \right)}}{{x + 1}}} \right] = 2 \cr & \Rightarrow {\log _5}x = 2 \cr & \Rightarrow x = {5^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 25 \cr}$

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\begin{aligned} \log_{5} (x^{2} + x) - \log_{5} (x + 1) = 2 \\ \Rightarrow \log_{5} (\frac{x^{2} + x}{x + 1}) = 2 \\ \Rightarrow \log_{5} [\frac{x (x + 1)}{x + 1}] = 2 \\ \Rightarrow \log_{5} x = 2 \\ \Rightarrow x = 5^{2} \\ = 25 \end{aligned}

\begin{aligned} \log_{5} (x^{2} + x) - \log_{5} (x + 1) = 2 \\ \Rightarrow \log_{5} (\frac{x^{2} + x}{x + 1}) = 2 \\ \Rightarrow \log_{5} [\frac{x (x + 1)}{x + 1}] = 2 \\ \Rightarrow \log_{5} x = 2 \\ \Rightarrow x = 5^{2} \\ = 25 \end{aligned}

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$\begin{aligned} If \log_{5} (x^{2} + x) - \log_{5} (x + 1) = 2, \\ then the value of x is - \end{aligned}$ $\eqalign{ & {\text{If }}{\log _5}\left( {{x^2} + x} \right) - {\log _5}\left( {x + 1} \right) = 2, \cr & {\text{then the value of }}x{\text{ is - }} \cr}$

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Hello Guest

If log5(x2+x)−log5(x+1)=2,then the value of x is - If log5(x2+x)−log5(x+1)=2,then the value of x is - \eqalign{ & {\text{If }}{\log _5}\left( {{x^2} + x} \right) - {\log _5}\left( {x + 1} \right) = 2, \cr & {\text{then the value of }}x{\text{ is - }} \cr}

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$\begin{aligned} If \log_{5} (x^{2} + x) - \log_{5} (x + 1) = 2, \\ then the value of x is - \end{aligned}$ $\eqalign{ & {\text{If }}{\log _5}\left( {{x^2} + x} \right) - {\log _5}\left( {x + 1} \right) = 2, \cr & {\text{then the value of }}x{\text{ is - }} \cr}$

NCERT
6 to 12