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kifflom [539]
2 years ago
10

Solve for the variable! 4 questions! Need to turn it in soon so ASAP!

Mathematics
1 answer:
d1i1m1o1n [39]2 years ago
6 0

Answer:

Step-by-step explanation:

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Solve for x<br><img src="https://tex.z-dn.net/?f=0%20%3D%20%20-%20%7B%20e%20%7D%5E%7B%20-%20x%7D%20%20%2B%203%20%7Be%7D%5E%7B3x%
Brilliant_brown [7]

\bf \textit{Logarithm Cancellation Rules} \\\\ \stackrel{\stackrel{\textit{we'll use this one}}{\downarrow }}{log_a a^x = x}\qquad \qquad a^{log_a x}=x~\hfill\stackrel{recall}{ln=log_e}\qquad log_e(e^z)=z \\\\[-0.35em] \rule{34em}{0.25pt}


\bf 0=-e^{-x}+3e^{3x}\implies e^{-x}=3e^{3x}\implies \cfrac{1}{e^x}=3e^{3x}\implies 1=e^x\cdot 3e^{3x} \\\\\\ 1=3e^xe^{3x}\implies \cfrac{1}{3}=e^{x+3x}\implies \cfrac{1}{3}=e^{4x}\implies ln\left( \cfrac{1}{3} \right)=ln\left( e^{4x} \right) \\\\\\ ln\left( \cfrac{1}{3} \right)=4x\implies \cfrac{ln\left( \frac{1}{3} \right)}{4}=x


and you plug that in your calculator to get about -0.27465307216702742285.

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