Question

Find the value of x which satisfies the following equation. log2(x−1)+log2(x+5)=4

Answer 1

$\quad \huge \quad \quad \boxed{ \tt \:Answer }$

$\qquad \tt \rightarrow \: x = 3$

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$\large \tt Solution \: :$

$\qquad \tt \rightarrow \: log_{2}(x - 1) log_{2}(x + 5) = 4$

$\qquad \tt \rightarrow \: log_{2} \{(x - 1)(x + 5) \} = 4$

[ log (x) + log (y) = log (xy) ]

$\qquad \tt \rightarrow \: ( x - 1)(x + 5) = {2}^{4}$

$\qquad \tt \rightarrow \: {x}^{2} + 5x - x - 5 = 16$

$\qquad \tt \rightarrow \: {x}^{2} + 4x - 5 - 16 = 0$

$\qquad \tt \rightarrow \: {x}^{2} + 4x -21 = 0$

$\qquad \tt \rightarrow \: {x}^{2} + 7x - 3x - 21 = 0$

$\qquad \tt \rightarrow \: x(x + 7) - 3(x + 7) = 0$

$\qquad \tt \rightarrow \: (x + 7)(x - 3) = 0$

$\qquad \tt \rightarrow \: x = - 7 \: \: or \: \: x = 3$

The only possible value of x is 3, since we can't operate logarithm with a negative integer in it.

$\qquad \tt \rightarrow \: x = 3$

Answered by : ❝ AǫᴜᴀWɪᴢ ❞