If the root of 4 log 10(10 x)-6 log 10 x=2 ⋅ 3 log 10(100 x2) is x=(p/q), where p q are coprime natural numbers then p + q =

Question

If the root of 4 log 10(10 x)-6 log 10 x=2 ⋅ 3 log 10(100 x2) is x=(p/q), where p q are coprime natural numbers then p + q = (A) 99 (B) 100 (C) 101 (D) 102

Tardigrade Admin · Accepted Answer

a=2 log x text and b=3 log x ∴ 4 a2-a b-18 b2=0 ∴ (a/b)=((2/3)) log x=(9/4)=((2/3))-2 log 10 x=-2 arrow x=(1/100)

Q. If the root of $4^{l o g_{10} (10 x)} - 6^{l o g_{10} x} = 2 \cdot 3^{l o g_{10} (100 x^{2})}$ is $x = \frac{p}{q}$ , where $p & q$ are coprime natural numbers then $p + q =$

99

100

101

102

$a = 2^{l o g x} and b = 3^{l o g x}$
$∴ 4 a^{2} - ab - 18 b^{2} = 0$
$∴ \frac{a}{b} = (\frac{2}{3})^{l o g x} = \frac{9}{4} = (\frac{2}{3})^{- 2}$
$lo g_{10} x = - 2 \to x = \frac{1}{100}$

Q. If the root of 4log10​(10x)−6log10​x=2⋅3log10​(100x2) is x=qp​, where p&q are coprime natural numbers then p+q=

99

100

101

102

a=2logx and b=3logx ∴4a2−ab−18b2=0 ∴ba​=(32​)logx=49​=(32​)−2 log10​x=−2→x=1001​

Q. If the root of $4^{l o g_{10} (10 x)} - 6^{l o g_{10} x} = 2 \cdot 3^{l o g_{10} (100 x^{2})}$ is $x = \frac{p}{q}$ , where $p & q$ are coprime natural numbers then $p + q =$

$a = 2^{l o g x} and b = 3^{l o g x}$
$∴ 4 a^{2} - ab - 18 b^{2} = 0$
$∴ \frac{a}{b} = (\frac{2}{3})^{l o g x} = \frac{9}{4} = (\frac{2}{3})^{- 2}$
$lo g_{10} x = - 2 \to x = \frac{1}{100}$