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Given a logarithm with the form log b M , use the change-of-base formula to rewrite it as a quotient of logs with any positive base n , where n 1.

  1. Determine the new base n , remembering that the common log, log ( x ) , has base 10, and the natural log, ln ( x ) , has base e .
  2. Rewrite the log as a quotient using the change-of-base formula
    • The numerator of the quotient will be a logarithm with base n and argument M .
    • The denominator of the quotient will be a logarithm with base n and argument b .

Changing logarithmic expressions to expressions involving only natural logs

Change log 5 3 to a quotient of natural logarithms.

Because we will be expressing log 5 3 as a quotient of natural logarithms, the new base, n = e .

We rewrite the log as a quotient using the change-of-base formula. The numerator of the quotient will be the natural log with argument 3. The denominator of the quotient will be the natural log with argument 5.

log b M = ln M ln b log 5 3 = ln 3 ln 5
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Change log 0.5 8 to a quotient of natural logarithms.

ln 8 ln 0.5

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Can we change common logarithms to natural logarithms?

Yes. Remember that log 9 means log 10 9 . So, log 9 = ln 9 ln 10 .

Using the change-of-base formula with a calculator

Evaluate log 2 ( 10 ) using the change-of-base formula with a calculator.

According to the change-of-base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm, which is the log base e .

log 2 10 = ln 10 ln 2 Apply the change of base formula using base  e . 3.3219 Use a calculator to evaluate to 4 decimal places .
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Evaluate log 5 ( 100 ) using the change-of-base formula.

ln 100 ln 5 4.6051 1.6094 = 2.861

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Access these online resources for additional instruction and practice with laws of logarithms.

Key equations

The Product Rule for Logarithms log b ( M N ) = log b ( M ) + log b ( N )
The Quotient Rule for Logarithms log b ( M N ) = log b M log b N
The Power Rule for Logarithms log b ( M n ) = n log b M
The Change-of-Base Formula log b M = log n M log n b           n > 0 , n 1 , b 1

Key concepts

  • We can use the product rule of logarithms to rewrite the log of a product as a sum of logarithms. See [link] .
  • We can use the quotient rule of logarithms to rewrite the log of a quotient as a difference of logarithms. See [link] .
  • We can use the power rule for logarithms to rewrite the log of a power as the product of the exponent and the log of its base. See [link] , [link] , and [link] .
  • We can use the product rule, the quotient rule, and the power rule together to combine or expand a logarithm with a complex input. See [link] , [link] , and [link] .
  • The rules of logarithms can also be used to condense sums, differences, and products with the same base as a single logarithm. See [link] , [link] , [link] , and [link] .
  • We can convert a logarithm with any base to a quotient of logarithms with any other base using the change-of-base formula. See [link] .
  • The change-of-base formula is often used to rewrite a logarithm with a base other than 10 and e as the quotient of natural or common logs. That way a calculator can be used to evaluate. See [link] .

Section exercises

Verbal

How does the power rule for logarithms help when solving logarithms with the form log b ( x n ) ?

Any root expression can be rewritten as an expression with a rational exponent so that the power rule can be applied, making the logarithm easier to calculate. Thus, log b ( x 1 n ) = 1 n log b ( x ) .

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What does the change-of-base formula do? Why is it useful when using a calculator?

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Algebraic

For the following exercises, expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.

log b ( 7 x 2 y )

log b ( 2 ) + log b ( 7 ) + log b ( x ) + log b ( y )

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log b ( 13 17 )

log b ( 13 ) log b ( 17 )

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ln ( 1 4 k )

k ln ( 4 )

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For the following exercises, condense to a single logarithm if possible.

ln ( 7 ) + ln ( x ) + ln ( y )

ln ( 7 x y )

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log 3 ( 2 ) + log 3 ( a ) + log 3 ( 11 ) + log 3 ( b )

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log b ( 28 ) log b ( 7 )

log b ( 4 )

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ln ( a ) ln ( d ) ln ( c )

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log b ( 1 7 )

log b ( 7 )

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For the following exercises, use the properties of logarithms to expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.

log ( x 15 y 13 z 19 )

15 log ( x ) + 13 log ( y ) 19 log ( z )

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ln ( a −2 b −4 c 5 )

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log ( x 3 y 4 )

3 2 log ( x ) 2 log ( y )

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log ( x 2 y 3 x 2 y 5 3 )

8 3 log ( x ) + 14 3 log ( y )

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For the following exercises, condense each expression to a single logarithm using the properties of logarithms.

log ( 2 x 4 ) + log ( 3 x 5 )

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ln ( 6 x 9 ) ln ( 3 x 2 )

ln ( 2 x 7 )

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2 log ( x ) + 3 log ( x + 1 )

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log ( x ) 1 2 log ( y ) + 3 log ( z )

log ( x z 3 y )

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4 log 7 ( c ) + log 7 ( a ) 3 + log 7 ( b ) 3

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For the following exercises, rewrite each expression as an equivalent ratio of logs using the indicated base.

log 7 ( 15 ) to base e

log 7 ( 15 ) = ln ( 15 ) ln ( 7 )

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log 14 ( 55.875 ) to base 10

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For the following exercises, suppose log 5 ( 6 ) = a and log 5 ( 11 ) = b . Use the change-of-base formula along with properties of logarithms to rewrite each expression in terms of a and b . Show the steps for solving.

log 11 ( 5 )

log 11 ( 5 ) = log 5 ( 5 ) log 5 ( 11 ) = 1 b

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log 11 ( 6 11 )

log 11 ( 6 11 ) = log 5 ( 6 11 ) log 5 ( 11 ) = log 5 ( 6 ) log 5 ( 11 ) log 5 ( 11 ) = a b b = a b 1

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Numeric

For the following exercises, use properties of logarithms to evaluate without using a calculator.

log 3 ( 1 9 ) 3 log 3 ( 3 )

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6 log 8 ( 2 ) + log 8 ( 64 ) 3 log 8 ( 4 )

3

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2 log 9 ( 3 ) 4 log 9 ( 3 ) + log 9 ( 1 729 )

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For the following exercises, use the change-of-base formula to evaluate each expression as a quotient of natural logs. Use a calculator to approximate each to five decimal places.

log 1 2 ( 4.7 )

2.23266

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Extensions

Use the product rule for logarithms to find all x values such that log 12 ( 2 x + 6 ) + log 12 ( x + 2 ) = 2. Show the steps for solving.

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Use the quotient rule for logarithms to find all x values such that log 6 ( x + 2 ) log 6 ( x 3 ) = 1. Show the steps for solving.

x = 4 ; By the quotient rule: log 6 ( x + 2 ) log 6 ( x 3 ) = log 6 ( x + 2 x 3 ) = 1.

Rewriting as an exponential equation and solving for x :

6 1 = x + 2 x 3 0 = x + 2 x 3 6 0 = x + 2 x 3 6 ( x 3 ) ( x 3 ) 0 = x + 2 6 x + 18 x 3 0 = x 4 x 3 x = 4

Checking, we find that log 6 ( 4 + 2 ) log 6 ( 4 3 ) = log 6 ( 6 ) log 6 ( 1 ) is defined, so x = 4.

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Can the power property of logarithms be derived from the power property of exponents using the equation b x = m ? If not, explain why. If so, show the derivation.

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Prove that log b ( n ) = 1 log n ( b ) for any positive integers b > 1 and n > 1.

Let b and n be positive integers greater than 1. Then, by the change-of-base formula, log b ( n ) = log n ( n ) log n ( b ) = 1 log n ( b ) .

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Does log 81 ( 2401 ) = log 3 ( 7 ) ? Verify the claim algebraically.

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Questions & Answers

what is math number
Tric Reply
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Sidiki Reply
Need help solving this problem (2/7)^-2
Simone Reply
x+2y-z=7
Sidiki
what is the coefficient of -4×
Mehri Reply
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
Alfred Reply
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
Kala Reply
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
Moses Reply
12, 17, 22.... 25th term
Alexandra Reply
12, 17, 22.... 25th term
Akash
College algebra is really hard?
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Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Adu
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
kinnecy Reply
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
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I rally confuse this number And equations too I need exactly help
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salma
Commplementary angles
Idrissa Reply
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im all ears I need to learn
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Uday
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greetings from Iran
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Nharnhar
Practice Key Terms 4

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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