





When (V^{P}) itself is raised to the Qth power^{} What is [(10^{2})]^{3} ? Obviously, we need multiply 10 by itself many times. Since[(10^{2})]^{3} should multiply the above by itself for a total of 3 times: in all 2.3=6 factors of 10. In other words, if 10 is raised to the second power, and the result is then raised to the 3rd power, that is the same as raising 10 to the (2.3) = 6th power. Indeed, the same argument can be made for any numbers, as long as they are raised to powers which are whole numbers. For any choice of whole numbers P and Q, if you raise some number V to power P and then raise the result to power Q, the result is like raising the number V to power (P.Q): Say V = 10^{P} so that P = log V. Raising V to power Q and keeping the same rules found for integers P and Q Taking the logarithm of both sides
In this manner, given a number V, logarithms help calculate V^{Q}, even if Q is not a whole number (as discussed below). The prescription: Take the numberV Find its logarithm P, Multiply P by Q to get Q log V. Say that is the number U. Find the number whose logarithm U is, that is, find 10^{U} . That is the value of V^{Q} . But how does one interpret, raising a number to the power of an arbitrary real number? (1) We may start with a simple example. Suppose If any powers a and b satisfy then V behaves the way one expects 10^{1/2} or 10^{0.5} to do. Namely, multiplying if by itself: (2) Similarly for the Qth root of 10the number which must be multiplied by itself Q times to get 10. If written as 10^{1/Q}, it would satisfy the equation for x^{(a+b)}, with 10^{1/Q} multiplied by itself Q times. (We do not go now into the question of how the Qth root is derived: there exist methods.) This can also be expressed using the earlier equation (in the box above). If 10^{1/Q} is raised to the Qth power and the earlier relation still holds, then Furthermore, if the general relation holds for any two factors, let Raising this to the Pth power This lets one formally define the power (P/Q), i.e. any rational number: first raise V to power 1/Q (that is, take the Qth root of V), then raise all that to the Pth power, by multiplying it by itself P times. (3) There is no good way to raise 10 to a power which is "irrational," a number which cannot be written as a fraction P/Q; for instance, raising 10 to power √2 or π . However, even though such numbers can never be exactly expressed as a fraction, there exist ways of approaching them through a series of fractions F1, F2, F3 ... which approximate them ever more closely; log π, for instance, is near 10^{22/7} and even closer to 10^{355/113}, and by continuing this sequence (or more conventionally, using a sequence of decimal fractions), one can approximate it as closely as one wishes. In general, the numbers N1, N2, N3...equal to 10 raised to the powers F1, F2, F3 approaching the irrational number will also get closer to each other. One guesses that if the process is taken far enough, the result represents 10 to the irrational power as accurately as one may wish. =========================== One example of raising numbers to power 3/2, which also shows a graphical application of logarithms, is in Keplers third law. Suppose we seek to examine that law. We have the list of average distances r of planets from the Sun and of corresponding orbital periods T , and they both grow together, though not in strict proportion. We suspect that T is some power of rbut how can we check this, and how can we find what that power is? The data (see section on Kepler's laws
We already know of course that Kepler found T^{2} was proportional to R^{3} (so columns 4 and 5 should be equal, except the values used here are not accurate), Kepler's famous 3rd law. It can be written Exploring FurtherOther fractional exponents arise from the gas laws. You may have learned that in a gas the pressure P of a given quantity of gas (say, one gram) is inversely proportional to its volume VThat, however, only holds true if the temperature stays the same. In fact, when you pump gas into a container of half the volume, it not only generates higher pressure, but you invest energy overcoming the pressure of the gas already there. As a result (as users of bicycle pumps know well) the gas heats up, and its pressure increases more than twice. It turns out that if heat is not allowed to flow in or out, a good approximation for the gas law is 
Author and Curator: Dr. David P. Stern
Mail to Dr.Stern: stargaze("at" symbol)phy6.org .
Last updated 10 November 2007
Curators: Robert Candey, Alex Young, Tamara Kovalick
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