# 0.9 Logarithms

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At t =0, the switch is closed. Because the voltage across the capacitor cannot change instantaneously, the value of the voltage across the resistor immediately after the closing of the swith ( v (0 + ) will be 10 V. After the switch has been closed, current will flow throughout the circuit and the voltage across the resistor will diminish exponentially. Because the decay can be described by the use of an exponential function, this is an example of exponential decay .

We can write an expression for the transient response, that is, the voltage across the resistor for positive time

$v\left(t\right)=\text{10}{e}^{-t/\text{RC}}$

Here, the units of v ( t ) are Volts.

The product of the resistance and the capacitance is called the time constant for the circuit and is often denoted as (τ). The time constant is measured in seconds. For this example, $\tau =\left(\text{100}×{\text{10}}^{3}\right)\left(8×{\text{10}}^{-6}\right)=\text{800}×{\text{10}}^{-3}=0\text{.}\text{800}s$

So the transient response of the circuit becomes

Question : At what instant of time will the transent response be equal to 5 Volts?

Solution : We have been asked to find the value of t that satisfies the following equation.

$5=\text{10}{e}^{-t/0\text{.}8}$

Let us divide each side of the equation by (10) and interchange the sides

${e}^{-t/0\text{.}8}=0\text{.}5$

We now take the natural logarithm of each side

$\frac{-t}{0\text{.}8}=-0\text{.}\text{693}$
$t=\left(0\text{.}8\right)\left(0\text{.}\text{693}\right)=0\text{.}\text{555}$

So we conclude that at the time 0.555 seconds after the switch closes, the value of the voltage across the resistor will be 5.0 volts.

## Compound interest

It is always essential to consider the financial aspects of an engineering project. The field of industrial engineering provides us with the analytical tools necessary to weigh the merits of competing designs.

Most situations surrounding the financial aspects of an engineering project involve the determination of what is most economical in the long run. That is, engineers must be aware of the costs and benefits of a project over a considerable period of time. In situations as these, it is important to consider the time value of money. Because of the existence of interest, the current value of a dollar is worth more than the value of a dollar some time in the future.

Interest can be defined as money that is paid for the use of money that has been borrowed. The rate of interest is the ratio between the interest chargeable or payable at the end of a period of time. This period of time is typically yearly, quarterly or monthly. In this module, we will restrict our attention to interest that is paid yearly or per annum .

As an example, a sum of money is invested at an annual rate of interest of 4%. One year later, the interest that would be paid on the investment would be $40 (4% of$1,000). So after one year, the initial sum would grow to a value of $1,040 one year later. Suppose that the$1,040 were invested for a second year at the end of the first year. At the end of the second year, the amount of interest that would be payable would be 4% of $1,040 or$41.60. The amount of interest earned in the second year exceeds the amount earned in the first year because of a phenomenon known as compound interest .

how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
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Tamia
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Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
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China
Cied
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I start with an easy one. carbon nanotubes woven into a long filament like a string
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many many of nanotubes
Porter
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what is the function of carbon nanotubes?
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what is system testing
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
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how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
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silver nanoparticles could handle the job?
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not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
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this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
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can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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