1108

• Run Multiple Choice Generator at h:\shares\physics
• Run Core Problems by giving the program 174 as your 'course'--enter this when asked for your 3-digit course number.  
• Do the 'quizzes' on Exponential Functions: Numbers and Exponential Functions: Graphs.
• If you miss a problem be sure to click on Comment and explain what you do and do not understand about the question and the given solution

• log( a * b) = log(a) + log(b)

Yes. log( a * b ) = log(a) + log(b).  This law is the inverse of the law of exponents x^(y + z) = x^y + x^z.

• log(a ^ b) = b * log(a)

Yes. log(a^b) = b log(a).  This law is the inverse of the law of exponents (x^y)^z = x^(yz).

• log( a * b) = log(a) * log(b)

No.  log( a * b ) = log(a) + log(b).  This law is the inverse of the law of exponents x^(y + z) = x^y + x^z.

• log(a ^ b) = log(a*b)

No.  log(a^b) = b log(a).  This law is the inverse of the law of exponents (x^y)^z = x^(yz).  Also log(a*b) = log(a) + log(b), as we saw earlier.

• log(a / b) = log(a) - log(b)

Yes. 

log(a / b) = log(a * b^-1)

= log(a) + log(b^-1)

= log(a) + (-1) log(b)

= log(a) - log(b).

• log(a) + log(b) = log(a * b)

Yes.  log( a * b ) = log(a) + log(b).  This law is the inverse of the law of exponents x^(y + z) = x^y + x^z.

• log(a / b) = log(a) / log(b)

No. 

log(a / b)

= log(a * b^-1)

= ...

= log(a) - log(b).

• log(a) + log(b) = log(a + b)

No.  log( a * b ) = log(a) + log(b).  This law is the inverse of the law of exponents x^(y + z) = x^y + x^z.

• b * log(a) = log(a ^ b).

Yes.  log(a^b) = b log(a).  This law is the inverse of the law of exponents (x^y)^z = x^(yz).

• log(a) - log(b) = log(a / b)

Yes.  b * log(a) = log(b *a).

• log(a^b) = b log(a). 

Yes.  This law is the inverse of the law of exponents (x^y)^z = x^(yz).

• log(a) - log(b) = log(a - b).

No.  log(a) - log(b) = log(a/b)

• log(20) = log(4) + log(5).

Yes.  Since 20 = 4 * 5, log(20) = log(4 * 5) = log(4) + log(5).

• log(1024) = 4 log(5)

No.  4 log(5) = log(5^4) = log(625), not log(1024).

• log(1024) = 5 log(4).

Yes.  5 log(4) = log(4^5) = log(1024).

• What two whole numbers is the log(7857) between?

On our table 7857 is between 1000 and 10,000 so the log is between 3 and 4.

The log of the number halfway between 1000 and 10,000 is more than halfway from 3 to 4, so the log of 7857 is clearly closer to 4 than to 3.

Solve the equation 5^(2x-4) = 79.

log(20) = log(4) + log(5)

log(1024) = 5 * log(4)

log(7284) is between

log(.00874) is between

y = log{base 10}(x) vs. x is inverse to y = 10^x vs. x.  The tables are inverse, the graphs are inverse.  You should be able to make a basic table and graph for both functions in 2 minutes or less.

Make a table of y = 10^x vs. x, then invert it to make a table of y = log{base 10}(x) vs. x. Sketch a quick graph of log{base 10}(x) vs. x.

What are log(100), log(.001) and log(10,000)?

log(100) = 2, easily read from the table.  This happens because 10^2 = 100, which should be clear from looking at both tables.

Similarly, log(.001) = -3 because 10^-3 = .001.

log(10,000) = 4, easily figured by extending the table.  This is so because 10^4 = 10,000.

Estimate log(50), log(-.3), log(.3) and log(7000).

50 lies between 10 and 100, so log(50) lies between log(10) = 1 and log(100) = 2.  We might guess about halfway:  maybe log(50) = 1.5.  This is a pretty good guess.

Looking at the concavity of the graph we see that the actual value is probably a little higher, or about 1.7.

Use your calculator to find log(50), log(-.3) and log(7000).

What is 10^2.4?  What is the log of this number?

What would you get if you took the log of both sides of the equation 10^(2x - 3) = .001?

We would get

• log(10^(2x-3)) = log(.001).

log(10^2.4) = 2.4, log(10^aardvark) = aardvark, and log(10^(2x-3) ) = 2x - 3.

log(.001) = -3, since 10^-3 = .001.

So our equation becomes

• 2x - 3 = -3.

Laws of exponents:

10^(a + b) = 10^a * 10^b

(10^a)^b = 10^(ab).

Laws of logarithms turn these laws around:

log(a * b) = log(a) + log(b)

log(a^b) = b * log(a).