If $p = l o g_{10} 20$ and $q = l o g_{10} 25$ . Find x such that $2 l o g_{10} (x + 1) = 2 p - q$

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Solution

The correct option is C 3
We have,
$p = l o g_{10} 20 \Rightarrow 10^{p} = 20$ .......(i)
$q = l o g_{10} 25 \Rightarrow 10^{q} = 25$ ........(ii)
Now,
$2 l o g_{10} (x + 1) = 2 p - q$
$\Rightarrow l o g_{10} (x + 1)^{2} = 2 p - q$
$\Rightarrow (x + 1)^{2} = 10^{2 p - q}$
$\Rightarrow (x + 1)^{2} = (10^{p})^{2} . 10^{- q}$
$\Rightarrow (x + 1)^{2} = \frac{20^{2}}{25}$
$\Rightarrow x^{2} + 1 + 2 x - 16 = 0$
$\Rightarrow x^{2} + 2 x - 15 = 0$
$\Rightarrow x^{2} + 5 x - 3 x - 15 = 0$
$\Rightarrow (x + 5) (x - 3) = 0$
Argument of log cannot be negative
So, x = 3
Hence, option c is correct.

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