The number of real solution(s) of the equation, ${log}_{10} (7 x - 9)^{2} + {log}_{10} (3 x - 4)^{2} = 2$ is/are

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Solution

The correct option is B $2$
${log}_{10} (7 x - 9)^{2} + {log}_{10} (3 x - 4)^{2} = 2$
$\Rightarrow {log}_{10} (7 x - 9)^{2} (3 x - 4)^{2} = 2 [∵ log a + log b = log (a b)]$
$\Rightarrow (7 x - 9)^{2} (3 x - 4)^{2} = 10^{2} = 100$
$\Rightarrow (21 x^{2} - 55 x + 36)^{2} = 100$
$\Rightarrow 21 x^{2} - 55 x + 36 = \pm 10$
$21 x^{2} - 55 x + 26 = 0$
$x = \frac{55 \pm \sqrt{3025 - 2184}}{42} = \frac{55 \pm 29}{42} = 2, \frac{13}{21}$
Only two real solution.
Ans: B

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The number of real solution(s) of the equation, log10(7x−9)2+log10(3x−4)2=2 is/are

The correct option is B 2log10(7x−9)2+log10(3x−4)2=2⇒log10(7x−9)2(3x−4)2=2[∵loga+logb=log(ab)]⇒(7x−9)2(3x−4)2=102=100⇒(21x2−55x+36)2=100⇒21x2−55x+36=±1021x2−55x+26=0x=55±√3025−218442=55±2942=2,1321Only two real solution.Ans: B

The number of real solution(s) of the equation, ${log}_{10} (7 x - 9)^{2} + {log}_{10} (3 x - 4)^{2} = 2$ is/are