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Solve ln ( x 3 ) 4 ln ( x ) = 1 .

x = 1 e

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When evaluating a logarithmic function with a calculator, you may have noticed that the only options are log 10 or log, called the common logarithm , or ln , which is the natural logarithm. However, exponential functions and logarithm functions can be expressed in terms of any desired base b . If you need to use a calculator to evaluate an expression with a different base, you can apply the change-of-base formulas first. Using this change of base, we typically write a given exponential or logarithmic function in terms of the natural exponential and natural logarithmic functions.

Rule: change-of-base formulas

Let a > 0 , b > 0 , and a 1 , b 1 .

  1. a x = b x log b a for any real number x .
    If b = e , this equation reduces to a x = e x log e a = e x ln a .
  2. log a x = log b x log b a for any real number x > 0 .
    If b = e , this equation reduces to log a x = ln x ln a .

Proof

For the first change-of-base formula, we begin by making use of the power property of logarithmic functions. We know that for any base b > 0 , b 1 , log b ( a x ) = x log b a . Therefore,

b log b ( a x ) = b x log b a .

In addition, we know that b x and log b ( x ) are inverse functions. Therefore,

b log b ( a x ) = a x .

Combining these last two equalities, we conclude that a x = b x log b a .

To prove the second property, we show that

( log b a ) · ( log a x ) = log b x .

Let u = log b a , v = log a x , and w = log b x . We will show that u · v = w . By the definition of logarithmic functions, we know that b u = a , a v = x , and b w = x . From the previous equations, we see that

b u v = ( b u ) v = a v = x = b w .

Therefore, b u v = b w . Since exponential functions are one-to-one, we can conclude that u · v = w .

Changing bases

Use a calculating utility to evaluate log 3 7 with the change-of-base formula presented earlier.

Use the second equation with a = 3 and e = 3 :

log 3 7 = ln 7 ln 3 1.77124 .

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Use the change-of-base formula and a calculating utility to evaluate log 4 6 .

1.29248

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Chapter opener: the richter scale for earthquakes

A photograph of an earthquake fault.
(credit: modification of work by Robb Hannawacker, NPS)

In 1935, Charles Richter developed a scale (now known as the Richter scale ) to measure the magnitude of an earthquake . The scale is a base-10 logarithmic scale, and it can be described as follows: Consider one earthquake with magnitude R 1 on the Richter scale and a second earthquake with magnitude R 2 on the Richter scale. Suppose R 1 > R 2 , which means the earthquake of magnitude R 1 is stronger, but how much stronger is it than the other earthquake? A way of measuring the intensity of an earthquake is by using a seismograph to measure the amplitude of the earthquake waves. If A 1 is the amplitude measured for the first earthquake and A 2 is the amplitude measured for the second earthquake, then the amplitudes and magnitudes of the two earthquakes satisfy the following equation:

R 1 R 2 = log 10 ( A 1 A 2 ) .

Consider an earthquake that measures 8 on the Richter scale and an earthquake that measures 7 on the Richter scale. Then,

8 7 = log 10 ( A 1 A 2 ) .

Therefore,

log 10 ( A 1 A 2 ) = 1 ,

which implies A 1 / A 2 = 10 or A 1 = 10 A 2 . Since A 1 is 10 times the size of A 2 , we say that the first earthquake is 10 times as intense as the second earthquake. On the other hand, if one earthquake measures 8 on the Richter scale and another measures 6, then the relative intensity of the two earthquakes satisfies the equation

log 10 ( A 1 A 2 ) = 8 6 = 2 .

Therefore, A 1 = 100 A 2 . That is, the first earthquake is 100 times more intense than the second earthquake.

How can we use logarithmic functions to compare the relative severity of the magnitude 9 earthquake in Japan in 2011 with the magnitude 7.3 earthquake in Haiti in 2010?

To compare the Japan and Haiti earthquakes, we can use an equation presented earlier:

9 7.3 = log 10 ( A 1 A 2 ) .

Therefore, A 1 / A 2 = 10 1.7 , and we conclude that the earthquake in Japan was approximately 50 times more intense than the earthquake in Haiti.

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Questions & Answers

f(x) = x-2 g(x) = 3x + 5 fog(x)? f(x)/g(x)
Naufal Reply
fog(x)= f(g(x)) = x-2 = 3x+5-2 = 3x+3 f(x)/g(x)= x-2/3x+5
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how to use fundamental theorem to solve exponential
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(mathematics) For a complex number a+bi, the principal square root of the sum of the squares of its real and imaginary parts, √a2+b2 . Denoted by | |. The absolute value |x| of a real number x is √x2 , which is equal to x if x is non-negative, and −x if x is negative.
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log tan (x/4+x/2)
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A Function F(X)=Sinx+cosx is odd or even?
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f(x)=1/1+x^2 |=[-3,1]
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You have downloaded the aplication Calculus Volume 1, tackling about lessons for (mostly) college freshmen, Calculus 1: Differential, and this group I think aims to let concerns and questions from students who want to clarify something about the subject. Well, this is what I guess so.
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CALCULUS
v=(x,y) وu=(x,y ) ∂u/∂x* ∂x/∂u +∂v/∂x*∂x/∂v=1
log tan (x/4+x/2)
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Practice Key Terms 7

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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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