**Review of Scientific Notation**

To express something in scientific notation you must have a number between 1 and 10, multiplied by some power of 10:

3000 = 3 x 10^{3}

15 = 1.5 x 10^{1}

0.0112 = 1.12 x 10^{-2}

To multiply two powers of 10 add the exponents; to divide, subtract the exponents:

10^{2}x 10^{1}= 10^{3}; [2 + 1 = 3]

10^{2}x 10^{2}= 10^{4}; [2 + 2 = 4]

10^{2}/ 10^{6}= 10^{-4}; [2 - 6 = -4]

10^{2}/ 10^{2}= 10^{0}= 1; [2 - 2 = 0]

You can also think of moving the decimal place to make a number between 1 and 10. If you need to move the decimal place two positions to the right, the power of ten is -2; 4 places to the left the power of ten is 4.

To multiply two numbers in scientific notation together, multiply the numbers between 1 and 10 first and then combine the powers of 10. Do something similar for division:

(2 x 10^{3}) x (3 x 10^{4}) = 6 x 10^{7}

(2 x 10^{3}) / (3 x 10^{4}) = 0.667 x 10^{-1}= 6.67 x 10^{-2}

(2 x 10^{3}) x (5 x 10^{4}) = 10 x 10^{7}= 1.0 x 10^{8}

(2 x 10^{3}) x (3 x 10^{4})^{2}= (2 x 10^{3}) x (9 x 10^{8}) = 1.8 x 10^{12}

**Logarithms**

The logarithm of a number is the power you need to apply to 10 to get that number. For example,

log(10^{4}) = 4.0

log(10^{8}) = 8.0

log(1) = log(10^{0}) = 0.0

log(0.01) = log(10^{-2}) = -2.0

So basically logarithms are just another way to order numbers. Note that negative logarithms mean that the numbers themselves are small, but greater than zero (like the 0.01 example above). For numbers that are not exact powers of 10, the logarithms fall between those of the powers of ten. For example, log(30) must fall between 1 (=log(10)) and 2 (=log(100)) because 30 lies between 10 and 100. The actual value of log(30) is 1.48.

A neat property of logarithms is shown in the next examples:

log(10^{8}x 10^{5}) = log(10^{13}) = 13 = 8 + 5 = log(10^{8}) + log(10^{5})

log(10^{3}x 10^{5}) = log(10^{8}) = 8 = 3 + 5 = log(10^{3}) + log(10^{5})

log(10^{-8}x 10^{5}) = log(10^{-3}) = -3 = -8 + 5 = log(10^{-8}) + log(10^{5})

So in general, log(ab) = log(a) + log(b). Likewise, log(a/b) = log(a) - log(b).**Units**

Units are often a source of error in homeworks and exams. Suppose we wish to convert 70 miles/hr into feet/second. First convert miles/hr into feet/hr, and then go on to feet/sec. One mile is 5280 feet, so if you travel at 1 mile/hour you cover 5280 feet/hr. Similarly, if you travel at 2 miles/hr, you cover 2x5280 = 10560 feet/hr. So 70 miles/hr = 70 x 5280 = 3.696 x 10^{5}feet/hr. But there are 3600 seconds per hour, and in one second you only cover 1/3600 of the distance you cover in an hour. Hence, 70 miles/hr = 70 x 5280 / 3600 = 102.7 feet/sec.

It is important to present the answer to a problem in units that are easily understood and can be compared to known scales. For example, the mass of a star may be quoted as 3.0 x 10^{33}gm, but this number is harder to compare with other stars than if we quote its equivalent, 1.5 M_{sun}(i.e., 1M_{sun}= 2.0 x 10^{33}gm.Practice converting the following to see if you get the idea:

1 km/s = 10^{5}cm/s

1 year = 3.15 x 10^{7}seconds

1 g/cm^{3}= 10^{3}kg/m^{3}

1 km/s = 0.98 parsecs/million years; where 1 pc = 3.09 x 10^{13}km

is a 'parsec'.**Velocity and Acceleration**

Velocity is the rate of change of distance with time, so its units are distance/time, such as km/s or miles/hr. Acceleration is the rate of change of velocity, so has units of velocity/time like (km/s)/s or km/s^{2}. Sometimes this is written as km s^{-2}. It is important to keep the units straight when trying to solve problems. For example, to find the distance x travelled in a time t for an object that accelerates from rest with acceleration a we could use

x = 0.5 a t^{2}

Notice how the units work out. If t is measured in seconds, and `a' in km s^{-2}, then the units on x = 0.5 a t^{2}are (km sec^{-2})(sec^{2}) = km. If we had used units of cm s^{-2}for a, then the answer for x would be in cm.