mozaMedical by Mozaik Education

The logarithmic identities

In view of the concept of logarithm let us examine which calculation rules help us calculate with logarithms.

We know that \latex{\log _{2} 32=5} and \latex{\log _{2} 4=2}, at the same time
\latex{\log _{2} (34\times 4)=\log _{2}128=7=5+2=\log _{2}32+\log _{2}4} .

The following theorem holds in general too.

THEOREM: The logarithm of a product is equal to the sum of the logarithms of the factors.

\latex{\log _{a}(x\times y)=\log _{a}x+\log _{a}y,} where \latex{x,y\gt 0;a\gt 0;a\neq 1}

Proof

Based on the definition let us express x, y and their product.
On the one hand
\latex{x=a^{\log_{a}}x^{} ;} \latex{y=a^{\log _{a}y^{} }}, thus

\latex{x\times y=a^{\log _{a}x^{} } \times a^{\log _{a}y^{} } = a^{\log _{a}x^{}+}x^{\log _{a}y^{} }.}

exponentiation identity I

On the other hand, from the definition

\latex{x\times y=a^{\log _{a}(x\times y)}.}

The two forms of the product – by using the strict monotonicity of the exponential function – imply:

\latex{\log _{a}(x\times y)=\log _{a} x+\log _{a}y.}

For example: \latex{\log _{6}12+\log _{6}18=\log _{6}(12\times 18)=\log _{6}216=3}.

⯁ ⯁ ⯁

Similarly to the previous example

\latex{\log _{2}(\frac{32}{4} )=\log _{2}8=3=5-2=\log _{2}32-\log _{2}4.}

In general:

THEOREM: The logarithm of a fraction is equal to the difference of the logarithms of the numerator and the denominator.

\latex{\log _{a}\frac{x}{y}=\log _{a}x-\log _{a}y,} where \latex{x,y\gt 0;a\gt 0;a\neq 1.}

Proof

By using the definition of logarithm: \latex{x=a^{\log _{a}x^{} }; y=a^{\log _{a}y^{} }.} Therefore

\latex{\frac{x}{y} =\frac{a^{\log _{a}x^{} } }{a^{\log _{a}y^{} } } =a^{\log _{a}x^{}-\log _{a}y^{} }.}

exponentiation identity II

On the other hand based on the definition:
\latex{\frac{x}{y} =a^{\log _{a}\frac{x}{y} ^{} } .}
The two forms of the fraction – by using the strict monotonicity of the exponential function – imply that

\latex{\log _{a}\frac{x}{y} =\log _{a}x^{} -\log _{a}y.}

For example: \latex{\log _{12}3-\log _{12}36=\log _{12}\frac{1}{12}=-1.}

⯁ ⯁ ⯁

Let us calculate:

\latex{\log _{2}32^{2}=\log _{2}1024=101=2\times 5=2\times \log _{2}32.}

The following theorem holds in general too:

THEOREM: The logarithm of a power is equal to the productof the exponent and the logarithm of the base.
\latex{\log _{a}x^{k}=k\times \log _{a}x,} where \latex{x\gt 0;a\gt 0;a\neq 1;k\in \R}.

Proof

Let us give the expressions by using the definition of logarithm.
On one hand \latex{x=a^{\log _{a}x^{} }}, therefore \latex{x^{k} =(a^{\log _{a}x^{} } )^{k}=a^{k\times \log _{a}x^{} }.}

On the other hand the definition implies:

\latex{x^{k}=a^{\log _{a}x^{k}} .}

The equality of the two powers – by using the strict monotonicity ofthe exponential function – implies that

\latex{\log _{a}x^{k}=k\times \log _{a}x.}

Note: Watch out! The logarithmic identities hold only if both sides have a meaning.
For example \latex{2\times \log x= \log x^{2}} holds only in the case of
\latex{x\gt 0.} In general, if \latex{x\neq 0}, then \latex{\log x^{2}=2\times \log |x|.}

Example 1

Let us calculate the values of the unknowns if

\latex{\log x=\log 13+\log 7;}

\latex{\log x=4\times \log 2-\log 5;}

\latex{\log _{3}x=\log _{3}11-2\times \log _{3}5+\log _{3}8.}

Solution

\latex{\log 13+\log 7=\log (13\times 7)=\log 91,} therefore \latex{x=91}
\latex{4\times \log 2-\log 5=\log 2^{4}-\log 5=\log \frac{2^{4} }{5} =\log \frac{16}{5},} therefore \latex{x=\frac{16}{5}}
\latex{\log _{3}11-2\times\log _{3}5+\log _{3}8=(\log _{3}11+\log _{3}8)-\log _{3}5^{2}=\log {3}\frac{11\times 8}{5^{2}}=\log _{3}\frac{88}{25},} therefore \latex{x= \frac{88}{25}}

Example 2

Let us calculate the values of the following expressions:

\latex{10^{\log 2+\log 3}}

\latex{6^{\log _{6}24-\log _{6}4^{} };}

\latex{5^{2\times \log _{5}3+3\times \log _{5}2};}

\latex{(\sqrt{40})^{4\times \log 8-\log 9}.}

Solution

\latex{10^{\log 2+\log 3}=10^{\log (2\times 3)}=10^{\log 6}=6}
\latex{6^{\log _{6}24-\log _{6}4}=6^{\log _{6}\frac{24}{4} } =6^{\log _{6}6}=6}
\latex{5^{2\times \log _{5}2}=5^{\log _{5}3^{2}+}\log _{5}2^{3}=5^{\log _{5}(3^{2}\times 2^{2}){} }=72}
\latex{(\sqrt{10})^{4\times \log 8- \log 9}=(10^{\frac{1}{2} } )^{4\times \log 8-\log 9}=10^{2\times\log 8-\frac{1}{2}\times \log 9}=10^{\log 8^{2}- \log 9^{\frac{1}{2}} }=10^{\log \frac{64}{3} } =\frac{64}{3}}

Pocket calculators and tables generally give the logarithm values of numbers with base \latex{ 10 } or e. How can we calculate a logarithm value with a different base?

THEOREM (BASE CHANGING FORMULA):
\latex{\log _{a}b=\frac{\log _{c}b}{\log _{c}a},} where \latex{a,b,c\gt 0;a\neq 1;c\neq 1}

Proof

Based on the definition of logarithm:

\latex{a^{\log _{a}b}=b.}

Let us express a as a power of \latex{ c }:

\latex{a=c^{\log _{c}a},}

and let us substitute it into the previous one:

\latex{(c^{\log _{c}a})^ {\log _{a}b}=b.}

Using the power of a power formula

\latex{c^{\log _{c}a\times \log _{a}b}=b.}

Due to the definition \latex{c^{\log _{c}b}=b,} so \latex{c^{\log _{c}a\times \log _{a}b}=c^{\log ^{c}b}.}

Due to the strict monotonicity of the exponential function

\latex{\log _{c}a\times \log _{a}b=\log _{c}b.}

Since

\latex{\log _{c}a\neq 0,} \latex{\log _{a}b=\frac{\log _{c}b}{\log _{c}a}.}

Example 3

What is the value of \latex{3^{\log _{9}16-\log _{27}8}} ?

Solution

Let us convert to the logarithm with base \latex{ 3 }:

\latex{\log _{9}16=\frac{\log _{3}16}{\log _{3}9}=\frac{\log _{3}16}{2}=\frac{1}{2}\times \log _{3}16.}

By using identity \latex{ III }:

\latex{\frac{1}{2}\times \log _{3}16=\log _{3}16^{\frac{1}{2}} =\log _{3}4.}

Let us act similarly with the logarithm with base \latex{ 27 }:

\latex{\log _{27}8=\frac{\log _{3}8}{\log _{3}27}=\frac{\log _{3}8}{3}=\frac{1}{3} \times \log _{3}8^{\frac{1}{3}} =\log _{3}2.}

Therefore

\latex{3^{\log _{9}16- \log _{27}8}=3^{\log _{3}4-\log _{3}2}=3^{\log _{3}2}=2.}

Exercises

{{exercise_number}}. Express x in terms of \latex{ a;\, b;\, c } and \latex{ d }:

\latex{\log x=\log a+\log b-\log c-\log d;}
\latex{\log x=2\times \log a-3\times \log b+0,5\times \log c-\frac{2}{3} \times \log d;}
\latex{\log x=-\log a-2\times \log b-3\times \log c-\log d;}
\latex{x=\log a-\log b-\log c-\log d.}

{{exercise_number}}. Calculate the values of the unknowns:

\latex{\log a=2\times \log 3+\log 11;}

\latex{\log b=\frac{1}{2} \times \log 4-\frac{3}{2} \times \log 9;}

\latex{\log c=\frac{2}{3} \times \log 8-\frac{1}{3} \times \log 27+2\times \log 5;}

\latex{\log d=\frac{1}{2} \times \log 15+\log 6-\frac{1}{2} \times \log 5+\frac{1}{2} \times \log 3.}

{{exercise_number}}. Calculate the values of the following powers:

\latex{10^{\log 7{}-\log 2};}

\latex{5^{\log _{5}2+\log _{5}7};}

\latex{3^{\log _{3}8+\log _{3}5- \log _{3}4};}

\latex{4^{\log _{4}3+\log _{2}9};}

\latex{5^{6\times \log _{125}2+\log _{25}9};}

\latex{2^{\log _{8}27-3\times \log _{64}9}.}

Puzzle

\latex{\log _{2}3\times \log _{3}4\times \log _{4}5\times \log _{5}6\times \log _{6}7\times \log _{7}8=?}

Mathematics 11.

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