Question

What is the true solution to the equation below? 2 in e in2×-in e in 10×= in 30 A x=30 B x=75 C x=150 D x=300

Answer 1

Answer:

Option B.

Step-by-step explanation:

Let as consider the given equation:

$2\ln e^{\ln 2x}-\ln e^{\ln 10x}=\ln 30$

It can be written as

$2(\ln 2x)-(\ln 10x)=\ln 30$         $[\because \ln e^a=a]$

$\ln (2x)^2-(\ln 10x)=\ln 30$        $[\because \ln a^b=b\ln a]$

$\ln \dfrac{4x^2}{10x}=\ln 30$        $[\because \ln \dfrac{a}{b}=\ln a-\ln b]$

$\ln \dfrac{2x}{5}=\ln 30$

On comparing both sides, we get

$\dfrac{2x}{5}=30$

Multiply both sides by 5.

$2x=150$

Divide both sides by 2.

$x=75$

Therefore, the correct option is B.