Question

What is the true solution to the equation below? 2 In e^In2x -In e^In10x = In 30

Answer 1

$The\ domain.\\D:x\in\mathbb{R^+}\\\\2\ln e^{\ln2x}-\ln e^{\ln10x}=\ln30\ \ \ |use\ \log_ab^n=n\log_ab\\\\2\ln2x\ln e-\ln10x\ln e=\ln30\ \ \ \ |use\ \log_aa=1\\\\2\ln2x-\ln10x=\ln30\ \ \ |use\ n\log_ab^n=\log_ab^n\\\\\ln(2x)^2-\ln10x=\ln30\\\\\ln4x^2-\ln10x=\ln30\ \ \ |use\ \log_ab-\log_ac=\log_a\dfrac{b}{c}\\\\\ln\dfrac{4x^2}{10x}=\ln30\\\\\ln\dfrac{2x}{5}=\ln30\iff\dfrac{2x}{5}=30\ \ \ \ |\cdot5\\\\2x=150\ \ \ \ |:2\\\\x=75\in D$

Answer: x = 75

Answer 2

Answer:

75

Step-by-step explanation: