Log (3x+1) - Log (2x+5)= 1

Mathematics

1 answer:

sleet_krkn [62]3 years ago

6 0

Answer:

$\huge \red { \boxed{x = - \frac{49}{17}}}$

Step-by-step explanation:

$log(3x + 1) - log(2x + 5) = 1 \\ \\ log \: \frac{3x + 1}{2x + 5} = log \: 10..( \because \: log \: 10 = 1) \\ \\ \frac{3x + 1}{2x + 5} = 10 \\ \\ 3x + 1 = 20x + 50 \\ 3x - 20x = 50 - 1 \\ \\ - 17x = 49 \\ \\ x = \frac{49}{ - 17} \\ \\ \huge \purple{ \boxed{x = - \frac{49}{17}}} \\$

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If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,

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<h2>Hello!</h2>

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The first option, the amount dumped after 5 days is 0.166 tons.

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

$\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)$

$(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166$