Ears are amazing.
The average human ear can just as easily recognize the whisper of the person in the next chair as the roar of a jet plane on takeoff. Most people know that, but many people do not realize that the difference in power between the two is approximately 10,000,000,000 times!
If you look up the decibel ratings of these two sounds, you will find that the quietest whispers are apx 10 db, and that a 747 on takeoff can hit 120 db.
That's only a difference of 100 db, how can that be the same as 10 BILLION times?
Every exponential expression can be written in logarithmic form. For example, the equation x = 2y is written as follows: y = log2x. In general, the equation logbn = a is equivalent to the equation ba = n. That is, b is the base, a is the exponent, and n is the power, or the result you obtain by raising b to the power of a. Notice that the exponential form of an expression emphasizes the power, while the logarithmic form emphasizes the exponent. More simply put, a logarithm (or “log” for short) is an exponent.
Perhaps the most common example of a logarithm is the Richter scale, which measures the magnitude of an earthquake. The magnitude is actually the logarithm base 10 of the amplitude of the quake. That is, m = log10A . This means that, for example, an earthquake of magnitude 4 is 10 times as strong as an earthquake with magnitude 3. We can see why this is true of we look at the logarithmic and exponential forms of the expressions: An earthquake of magnitude 3 means 3 = log10A. The exponential form of this expression is 103 = A. Thus the amplitude of the quake is 1,000. Similarly, a quake with magnitude 4 has amplitude 104 = 10,000.
Solving Logarithmic Equations
In general, to solve an equation means to find the value(s) of the variable that makes the equation a true statement. To solve log equations, we have to think about what “log” means.
Consider the equation log2x = 5 . What is the exponential form of this equation?
The equation log2x = 5 means that 25 = x . So the solution to the equation is x = 25 = 32.
In some log equations, both sides of the equation contain a log. To solve these equations, use the following rule: logb f(x) = logb g(x) → f(x) = g(x).
In other words, set the bases equal and solve for the variable in the exponent by treating the exponents on both sides of the equation as simple polynomials.
Rewrite each exponential expression as a log expression.
In order to rewrite an expression, you must identify its base, its exponent, and its power.
- 34 = 81
The 3 is the base, so it is placed as the subscript in the log expression. The 81 is the power, and so it is placed after the “log”. Thus we have: 34 = 81 is the same as log381 = 4. To read this expression, we say “the logarithm base 3 of 81 equals 4.” This is equivalent to saying “ 3 to the 4th power equals 81.”
- b4x = 52
The b is the base, and the expression 4x is the exponent, so we have: logb52 = 4x . We say, “log base b of 52, equals 4x.”
Evaluate the function f(x) = log2x for the following values.
To determine the value of log22, you can ask yourself: “2 to what power equals 2?” Answering this question is often easy if you consider the exponential form: 2? = 2.
The missing exponent is 1. So we have f(2) = log22 = 1.
The missing exponent is 0. So we have f(1) = log21 = 0.
There is no such exponent. Therefore f(-2) = log2 -2 does not exist.
Solve the equation log2(3x-1) = log2(5x - 7) for x.
Because the logarithms have the same base (2), the arguments of the log (the expressions 3x - 1 and 5x - 7) must be equal. So we can solve as follows:
|log2(3x-1)||= log2(5x - 7)|
|3x - 1||= 5x - 7|
|3x + 6||= 5x|
Rewrite the following logarithmic expressions in exponential form.
- log10 100 = 2
The base is 10, and the exponent is 2, so we have 102 = 100.
- logb w = 5
The base is b, and the exponent is 5, so we have b5 = w.
Solve each equation for
- log4 x = 3
Writing the equation in exponential form gives us the solution: x = 43 = 64.
- log5 (x + 1) = 2
Writing the equation in exponential form gives us a new equation: 52 = x + 1.
We can solve this equation for x:
|52||= x + 1|
|25||= x + 1|
- 1 + 2log3 (x - 5) = 7
First, we have to isolate the log expression:
|1 + 2log3(x - 5)||= 7|
|2log3(x - 5)||= 6|
|log3(x - 5)||= 3|
Now, we can solve the equation by rewriting it in exponential form:
|log3(x - 5)||= 3|
|33||= x - 5|
|27||= x - 5|
- State the expression in English:
- Write the given equation in logarithmic form:
- Determine if the two equations below are equivalent and label them according to function family:
Evaluate the logarithms:
log4625 log664 log3216
Evaluate each logarithm at the indicated value of x.
f(x)=log2xfor x=32 f(x)=log3xfor x=1 f(x)=log4xfor x=2 f(x)=log10xfor x=1100
Solve for x:
log3(2x+2)=log3(x−4) log7(x3)=log7(x−4) log48x−−√3=log48 log5(x+9)=log53(x−2)
To see the Review answers, open this PDF file and look for section 3.5.