Precalculus I Class 03/27


We note that a formula can 'look right' but be very wrong.  Be sure to always check out your operations on logarithms with the laws.

To see why log(ab) = log(a) + log(b), first note that 10(x+y) = 10^x * 10^y, from which it follows that log(10^(x+y) ) = log(10^x * 10^y) = log(10^x) + log(10^y) = x + y.

Whatever a and b are, each can be expressed as a power of 10.  If a = 10^x and b = 10^y, which is the case for some x and some y, then log(ab) = x + y.  But since a = 10^x, log a = x and since b = 10^y, log b = y, so x + y = log a + log b.  It follows that log(ab) = log(a) + log(b).

Note that log{base b}(x) is the power of b required to get x.

 

Find these logs:

log{base 2}(16) is the power of 2 we need to get 16.  

log{base 3}(81) is the power of 3 we need to get 81.  

log{base 5}(125) is the power of 5 we need to get 125.  

log{base 4}(64) is the power of 4 we need to get 64.  

log{base 4}(32).  4^2 = 16 and 4^3 = 64.  So log{base 4}(32) is between 2 and 3.  

log{base 8}(32).  We need the power of 8 that is equal to 32.  

Suggested problems:  

log{base 25}(125)

log{base 9}(243)

log{base 16}(32)

log{base 27}(243)

log{base 2}(16) is 4.  Verify using your calculator and the identity log{base b}(x) = log x / log b:  

log{base 3}(81) is 4.  Verify using your calculator and the identity log{base b}(x) = log x / log b:  

log{base 5}(125) is 3.  Verify using your calculator and the identity log{base b}(x) = log x / log b:  

log{base 4}(64) is 3.  Verify using your calculator and the identity log{base b}(x) = log x / log b:  

log{base 4}(32) is 2.5.

log{base 8}(32).  

 

We solve the equation 5^(7x-3) = 14.

The equation 2^(3x - 4) + 2^(5x - 1) = 7 looks like it should be amenable to the same treatment as in the preceding problem.  Taking logs of both sides looks like it should bring the x terms out of the numerator.  However there is a serious pitfall in the way of this strategy.

The problem is that log(a+b) is not equal to log a + log b, as we saw earlier.  So if after writing 

log [2^(3x - 4) + 2^(5x - 1)] = log(7)

we try to say that log(2^(3x-4)) + log(2^(5x - 1) ) = log(7), we are falling into the fallacy of linearity described earlier.

It turns out that this problem cannot be solved to find an exact value of x.  We can always approximate the solution, but there is no way to find an exact solution.

The equation 2^(1/x) = 7 * 2^(x+2) can be solved by taking the log of both sides.  

The problem is continues on the next figure.

 

Be sure to understand that log 2 and log 7 are just numbers, which we can easily evaluate using calculators.

We can now see that our equation is quadratic in x, containing x^2, x and constant terms.  We will put this equation into the form a x^2 + b x + b and then we will apply the quadratic formula so find values of x which satisfy the equation.

 

 

Suppose that in linearizing a data set we find that log(y) vs. log(x) gives us a straight line with slope .037 and vertical intercept 2.4.  What is the function that models our data?

To solve the equation log(y) = .037 log(x) + 2.4 we first write the equation 10^left side = 10^right side, as shown in the second line. 

We finally see that the data is modeled by the equation y = 251 x^.037, which is the power function x^p, with p = .037, vertically stretched by factor 251.