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Gekata [30.6K]
4 years ago
15

Solve log_8 12 = x – 2 . Round your answer to four decimal places.

Mathematics
2 answers:
Aleksandr-060686 [28]4 years ago
3 0

Step-by-step answer:

Let's attack the left side:

log_8 12 = log (base 8) 12 = log(12)/log(8) = 1.1949875 (approx.)

Next solve

log_8 12 = x-2

log(12)/log(8) = x-2

x = 2 + log(12)/log(8)

= 3.195 (to one thousandth)

Zina [86]4 years ago
3 0

For this case we have the following expression:

log_ {8} (12) = x-2

We rewrite how:

x-2 = log_ {8} (12)

We add 2 to both sides of the equation:

x = log_ {8} (12) +2

We rewrite log_ {8} (12) aslog_ {8} (2 ^ 2 * 3):

x = log_ {8} (2 ^ 2 * 3) +2

We apply logarithm properties:

x = log_ {8} (2 ^ 2) + log_ {8} (3) +2

We apply logarithm properties:

x = 2log_ {8} (2) + log_ {8} (3) +2

The logarithmic base 8 of 2 is \frac {1} {3}:

x = 2 (\frac {1} {3}) + log_ {8} (3) +2\\x = \frac {2} {3} + log_ {8} (3) +2

We simplify:

x = \frac {2} {3} + 2 + log_ {8} (3)\\x = \frac {2 + 6} {3} + log_ {8} (3)\\x = \frac {8} {3} + log_ {8} (3)

Decimal form:

3,195

Answer:

x = \frac {8} {3} + log_ {8} (3) = 3,195

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The population of mosquitoes in a certain area increases at a rate proportional to the current population, and in the absence of
Alexandra [31]

Answer:

Population of mosquitoes in the area at any time t is:

P(t) =504,943.26  -104,943.26e^{0.693t}

Step-by-step explanation:

assume population at any time t = P(t)

population increases at a rate proportional to the current population:

⇒dP/dt ∝ P

 \implies \frac{dP}{dt} =kP----(1)

where k is constant rate at which population is doubled

solving (1)

ln|P(t)|=kt +C\\P(t)= e^{kt+C}\\P(t)=Ce^{kt}

t=0\\P(0) = P_{o}\\\implies C= P_{o}\\P(t) =P_{o}e^{kt}\\ ---- (2)

initial population = 400,000

population is doubled every week

                                                 ⇒P(1)=2P(0)

Using (2)

                                 P_{o}e^{k(1)} = 2P_{o} e^{k(0)}\\

                                            e^{k} =2\\k=ln|2|\\

In presence of predators amount is decreased by 50,000 per day

Then amount decreased per week = 350,000

In this case (1) becomes

\frac{dP}{dt}=kP-350,000\\\frac{dP}{dt} - kP=-350,000\\ ---(3)

solving (3) by calculating integrating factor

                                          I.F=e^{\int-k dt}

Multiplying I.F with all terms of (3)

e^{-kt}\frac{dP}{dt} - ke^{-kt}P =-350,000 e^{-kt}\\\frac{d}{dt}(e^{-kt}P) =  -350,000 e^{-kt}

Integrating w.r.to t

                         e^{-kt}P(t)= \frac{350,000e^{-kt}}{k} +C

                         P(t) =\frac{350,000}{k} +Ce^{kt}\\

                                          k=ln|2| =0.693

                          P(t) =504,943.26 + Ce^{0.693t}\\

at t=0

                        P(0) =504,943.26 + Ce^{0.693(0)}

                        400,000 =504,943.26 + C

                           C = -104,943.26

So, population of mosquitoes in the area at any time t is

                  P(t) =504,943.26  -104,943.26e^{0.693t}

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For this case we have an equation of the form:
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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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