$The number of value(s) of x for which {log}_{2 x - 3} (4 x + 5) = 1 is$

1

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0

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Solution

The correct option is B $0$
Given: ${log}_{2 x - 3} (4 x + 5) = 1$
Now, for logarithm to be defined:
$2 x - 3 > 0 & 2 x - 3 \neq 1 \Rightarrow x > \frac{3}{2} & x \neq 2 \Rightarrow x \in (\frac{3}{2}, \infty) - {2} \dots (i)$
$& 4 x + 5 > 0 \Rightarrow x \in (\frac{- 5}{4}, \infty) \dots (i i)$
Thus, from both conditions, we get:
$x \in (\frac{3}{2}, \infty) - {2} \dots (A)$
Now, ${log}_{2 x - 3} (4 x + 5) = 1$
$\Rightarrow 2 x - 3 = 4 x + 5 \Rightarrow 2 x = - 8 \Rightarrow x = - 4$
But $x = - 4$ does not belong to the interval $A$
$\Rightarrow$ The expression has no real value of $x .$

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