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The exponential function f ( x ) = b x is one-to-one, with domain ( , ) and range ( 0 , ) . Therefore, it has an inverse function, called the logarithmic function with base b . For any b > 0 , b 1 , the logarithmic function with base b , denoted log b , has domain ( 0 , ) and range ( , ) , and satisfies

log b ( x ) = y if and only if b y = x .

For example,

log 2 ( 8 ) = 3 since 2 3 = 8 , log 10 ( 1 100 ) = −2 since 10 −2 = 1 10 2 = 1 100 , log b ( 1 ) = 0 since b 0 = 1 for any base b > 0 .

Furthermore, since y = log b ( x ) and y = b x are inverse functions,

log b ( b x ) = x and b log b ( x ) = x .

The most commonly used logarithmic function is the function log e . Since this function uses natural e as its base, it is called the natural logarithm    . Here we use the notation ln ( x ) or ln x to mean log e ( x ) . For example,

ln ( e ) = log e ( e ) = 1 , ln ( e 3 ) = log e ( e 3 ) = 3 , ln ( 1 ) = log e ( 1 ) = 0 .

Since the functions f ( x ) = e x and g ( x ) = ln ( x ) are inverses of each other,

ln ( e x ) = x and e ln x = x ,

and their graphs are symmetric about the line y = x ( [link] ).

An image of a graph. The x axis runs from -3 to 3 and the y axis runs from -3 to 4. The graph is of two functions. The first function is “f(x) = e to power of x”, an increasing curved function that starts slightly above the x axis. The y intercept is at the point (0, 1) and there is no x intercept. The second function is “f(x) = ln(x)”, an increasing curved function. The x intercept is at the point (1, 0) and there is no y intercept. A dotted line with label “y = x” is also plotted on the graph, to show that the functions are mirror images over this line.
The functions y = e x and y = ln ( x ) are inverses of each other, so their graphs are symmetric about the line y = x .

At this site you can see an example of a base-10 logarithmic scale.

In general, for any base b > 0 , b 1 , the function g ( x ) = log b ( x ) is symmetric about the line y = x with the function f ( x ) = b x . Using this fact and the graphs of the exponential functions, we graph functions log b for several values of b > 1 ( [link] ).

An image of a graph. The x axis runs from -3 to 3 and the y axis runs from 0 to 4. The graph is of three functions. All three functions a log functions that are increasing curved functions that start slightly to the right of the y axis and have an x intercept at (1, 0). The first function is “y = log base 10 (x)”, the second function is “f(x) = ln(x)”, and the third function is “y = log base 2 (x)”. The third function increases the most rapidly, the second function increases next most rapidly, and the third function increases the slowest.
Graphs of y = log b ( x ) are depicted for b = 2 , e , 10 .

Before solving some equations involving exponential and logarithmic functions, let’s review the basic properties of logarithms.

Rule: properties of logarithms

If a , b , c > 0 , b 1 , and r is any real number, then

1. log b ( a c ) = log b ( a ) + log b ( c ) (Product property) 2. log b ( a c ) = log b ( a ) log b ( c ) (Quotient property) 3. log b ( a r ) = r log b ( a ) (Power property)

Solving equations involving exponential functions

Solve each of the following equations for x .

  1. 5 x = 2
  2. e x + 6 e x = 5
  1. Applying the natural logarithm function to both sides of the equation, we have
    ln 5 x = ln 2 .

    Using the power property of logarithms,
    x ln 5 = ln 2 .

    Therefore, x = ln 2 / ln 5 .
  2. Multiplying both sides of the equation by e x , we arrive at the equation
    e 2 x + 6 = 5 e x .

    Rewriting this equation as
    e 2 x 5 e x + 6 = 0 ,

    we can then rewrite it as a quadratic equation in e x :
    ( e x ) 2 5 ( e x ) + 6 = 0 .

    Now we can solve the quadratic equation. Factoring this equation, we obtain
    ( e x 3 ) ( e x 2 ) = 0 .

    Therefore, the solutions satisfy e x = 3 and e x = 2 . Taking the natural logarithm of both sides gives us the solutions x = ln 3 , ln 2 .
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Solve e 2 x / ( 3 + e 2 x ) = 1 / 2 .

x = ln 3 2

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Solving equations involving logarithmic functions

Solve each of the following equations for x .

  1. ln ( 1 x ) = 4
  2. log 10 x + log 10 x = 2
  3. ln ( 2 x ) 3 ln ( x 2 ) = 0
  1. By the definition of the natural logarithm function,
    ln ( 1 x ) = 4 if and only if e 4 = 1 x .

    Therefore, the solution is x = 1 / e 4 .
  2. Using the product and power properties of logarithmic functions, rewrite the left-hand side of the equation as
    log 10 x + log 10 x = log 10 x x = log 10 x 3 / 2 = 3 2 log 10 x .

    Therefore, the equation can be rewritten as
    3 2 log 10 x = 2 or log 10 x = 4 3 .

    The solution is x = 10 4 / 3 = 10 10 3 .
  3. Using the power property of logarithmic functions, we can rewrite the equation as ln ( 2 x ) ln ( x 6 ) = 0 .
    Using the quotient property, this becomes
    ln ( 2 x 5 ) = 0 .

    Therefore, 2 / x 5 = 1 , which implies x = 2 5 . We should then check for any extraneous solutions.
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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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