To solve this problem it is necessary to apply the rules and concepts related to logarithmic operations.
From the definition of logarithm we know that,
In this way for the given example we have that a logarithm with base 10 expressed in the problem can be represented as,
We can express this also as,
By properties of the logarithms we know that the logarithm of a power of a number is equal to the product between the exponent of the power and the logarithm of the number.
So this can be expressed as
Since the definition of the base logarithm 10 of 10 is equal to 1 then
The value of the given logarithm is equal to 6
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Kepler noticed an imaginary line drawn from a planet to the Sun and this line swept out an equal area of space in equal times, If we then draw a triangle out from the Sun to a planet’s position at one point in time, it is notice that the area doesn't change even after the planet has left the original position say like after 2 to 3days or 2hours. So to have same area of triangle means that the the planet move faster when that are closer to the sun and slowly when they are far from the sun.
This led to Kepler's law of orbital motion.
First Law: Planetary orbits are elliptical with the sun at a focus.
Second Law: The radius vector from the sun to a planet sweeps equal areas in equal times.
Third Law: The ratio of the square of the period of revolution and the cube of the ellipse semi-major axis is the same for all planets.
It is this Kepler's law that makes Newton to come up with his own laws on how planet moves the way they do.
"<em>F = dP/dt. </em> The net force acting on an object is equal to the rate at which its momentum changes."
These days, we break up "the rate at which momentum changes" into its units, and then re-combine them in a slightly different way. So the way WE express and use the 2nd law of motion is
"<em>F = m·A.</em> The net force on an object is equal to the product of the object's mass and its acceleration."
The two statements say exactly the same thing. You can take either one and work out the other one from it, just by working with the units.
