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 10db, and that a 747 on takeoff can hit 120db.
That's only a difference of 100db, how can that be the same as 10 BILLION times?
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James Sousa: Solving Logarithmic Equations
Guidance
Every exponential expression can be written in logarithmic form. For example, the equation x = 2 ^{ y } is written as follows: y = log _{ 2 } x . In general, the equation log _{ b } n = a is equivalent to the equation b ^{ a } = 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 = log _{ 10 } A . 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 = log _{ 10 } A . The exponential form of this expression is 10 ^{ 3 } = A . Thus the amplitude of the quake is 1,000. Similarly, a quake with magnitude 4 has amplitude 10 ^{ 4 } = 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 log _{ 2 } x = 5 . What is the exponential form of this equation?

 The equation log _{ 2 } x = 5 means that 2 ^{ 5 } = x . So the solution to the equation is x = 2 ^{ 5 } = 32.
In some log equations, both sides of the equation contain a log. To solve these equations, use the following rule:

 log _{ b } f ( x ) = log _{ b } 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.
Example A
Rewrite each exponential expression as a log expression.

a. 3 ^{ 4 } = 81 b. b ^{ 4 } ^{ x } = 52
Solution:
 a. In order to rewrite an expression, you must identify its base, its exponent, and its power. 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: 3 ^{ 4 } = 81 is the same as log _{ 3 } 81 = 4 .

 To read this expression, we say “the logarithm base 3 of 81 equals 4.” This is equivalent to saying “ 3 to the 4 ^{ th } power equals 81.”
 b. The b is the base, and the expression 4x is the exponent, so we have: log _{ b } 52 = 4 x . We say, “log base b of 52, equals 4 x .”
Example B
Evaluate the function f ( x ) = log _{ 2 } x for the values:
a.) x = 2  b.) @$x = 1@$  c.) @$x = 2@$ 

Solution:
 a.) If @$x = 2@$ , we have:
f ( x ) = log _{ 2 } x  

f (2) = log _{ 2 } 2 

 To determine the value of log _{ 2 } 2, 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) = log _{ 2 } 2 = 1
 b.) If @$x = 1@$ , we have:
f ( x ) = log _{ 2 } x  

f (1) = log _{ 2 } 1 

 As we did in (a), we can consider the exponential form: 2 ^{ ? } = 1. The missing exponent is 0. So we have f (1) = log ^{ 2 } 1 = 0.
 c. If @$x = 2@$ , we have:
f ( x )  =log _{ 2 } x  

f (2)  = log _{ 2 } 2 

 Again, consider the exponential form: 2 ^{ ? } = 2. There is no such exponent. Therefore f (2) = log _{ 2 } 2 does not exist.
Example C
Consider the equation log _{ 2 } (3 x 1) = log _{ 2 } (5 x  7).
Because the logarithms have the same base (2), the arguments of the log (the expressions 3 x  1 and 5 x  7) must be equal . So we can solve as follows:
log _{ 2 } (3 x 1)  = log _{ 2 } (5 x  7)  

3 x  1  = 5 x  7  
+7 +7  
3 x + 6  = 5 x  
3 x  
6  = 2 x  
x  = 3 
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Guided Practice
1) Solve for x : log _{ 2 } (9 x ) = log _{ 2 } (3 x + 8)
2) Rewrite the logarithmic expressions in exponential form.

a. log _{ 10 } 100=2 b. log _{ b } w = 5
3) Solve each equation for @$x@$ :
a.) log _{ 4 } x = 3  b.) log _{ 5 } ( x + 1) = 2  c.) 1 + 2log _{ 3 } ( x  5) = 7 

4) Write the given equation in logarithmic form: @$5^2 = 25@$
Answers
1) The log equation implies that the expressions 9 x and 3 x + 8 are equal:
@$log_2(9x)@$  @$= log_2(3x + 8)@$  

@$9x@$  @$= 3x + 8@$  
@$3x@$  @$3x@$  
@$6x@$  @$= 8@$  
@$x@$  @$= \frac{8} {6}@$  
@$x@$  @$= \frac{4} {3}@$ 
2) a) The base is 10, and the exponent is 2, so we have: 10 ^{ 2 } = 100
 b) The base is b, and the exponent is 5, so we have: b ^{ 5 } = w .
3) a) Writing the equation in exponential form gives us the solution: x = 4 ^{ 3 } = 64.
 b) Writing the equation in exponential form gives us a new equation: 5 ^{ 2 } = x + 1.

 We can solve this equation for x :
5 ^{ 2 }  = x + 1  

25  = x + 1  
x  = 24 
 c) First we have to isolate the log expression:
1 + 2log _{ 3 } ( x  5)  = 7  

2log _{ 3 } ( x  5)  = 6  
log _{ 3 } (x  5)  = 3 
 Now we can solve the equation by rewriting it in exponential form:
log _{ 3 } (x  5)  = 3  

3 ^{ 3 }  = x  5  
27  = x  5  
x  = 32 
4) The exponential function is in the form: @$ y = b^x@$
 That means the corresponding logarithmic function is @$x = log _b y@$
In this problem, the base @$b@$ in the exponential function the same as the base @$b@$ in the logarithm, so we substitute.
Substitute 2 for @$x@$ , 5 for @$b@$ and 25 for @$y@$ in the logarithmic function.
 This gives: @$2 = log_5 25@$
Explore More
 State the expression in English: @$log_3 243@$
 Write the given equation in logarithmic form: @$(\frac{1}{3})^{5} = 243@$
 Determine if the two equations below are equivalent and label them according to function family: @$y = log_a x@$ and @$x = a^y@$
Evaluate the logarithms:
 @$log_4 625@$
 @$log_6 64@$
 @$log_3 216@$
Evaluate each logarithm at the indicated value of x.
 @$f(x) = log_2 x@$ for @$x = 32@$
 @$f(x) = log_3 x@$ for @$x = 1@$
 @$f(x) = log_4 x@$ for @$x = 2@$
 @$f(x) = log_{10} x@$ for @$x = \frac{1}{100}@$
Solve for x :
 @$log_3 (2x+2) = log_3 (x4)@$
 @$log_7 (\frac{x}{3}) = log_7 (x  4)@$
 @$log_4 \sqrt[3]{8x} = log_4 8@$
 @$log_5 (x + 9) = log_5 3(x  2)@$
 Evaluate @$log_4 x@$ for @$\frac{(x + 7)}{2} = 1204@$