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

Answer: x = 75

8 0

3 years ago

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Does anyone know what to do. It’s for 6th grade math. The pencil part is a guess that I did.

HACTEHA [7]

Hullo! this looks complicated i know, but it's actually quite simple. if 2 bars feed 6 aliens, that means every number on top will be one third of the number on the bottom. so it would go like this (I'll go up to 9, and it should make sense from there)

2 | 2.3 | 2.6 | 3
------------------------
6 | 7 | 8 | 9

i hope this helped! if all else fails, just ask for help. it really does go a long way

5 0

3 years ago