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Exponentials and Logarithms

April 29th, 2009 |

Author: Merrin

An exponential is a base carried to some exponent. Here b is the base and x is the exponent or power.

$b^{x}$

If x is a positive integer then there is a simple interpretation.

$b^{n} =\mathop{bb\cdot \cdot \cdot bb}\limits_{n\, times}$

If n is a negative integer then

$b^{-n} =\mathop{\cfrac{1}{b} \cfrac{1}{b} \cdot \cdot \cdot \cfrac{1}{b} \cfrac{1}{b} }\limits_{n\, times}$

If n is a fraction or irrational then exponentiation then you have to resort to some more advanced reasoning.

\includegraphics[width= 5in]{../images/Logs.pdf} \caption{Graphs of logarithms with some common bases} \includegraphics[width= 5in]{../images/plotexponentials.pdf} \caption{Graphs of common exponential functions} The logarithm is the inverse of exponentiation.

${\log _{b} b^{x} =x\log _{b} b=x}$

${b^{\log _{b} x} =x}$

There are a number of laws for exponents and logarithms used in their manipulation.

${(ab)^{n} =a^{n} b^{n} }$

${\left(\cfrac{a}{b} \right)^{n} =\cfrac{a^{n} }{b^{n} } }$

${(a^{n} )^{m} =a^{nm} } \\ {a^{n} a^{m} =a^{n+m} }$

There are also some useful logarithm laws.

${\log _{b} (MN)=\log _{b} M+\log _{b} N}$

${\log (M/N)=\log _{b} M-\log _{b} N}$

${\log _{b} b=1\quad b\ne 1 b>0}$

${\log _{a} b\log _{b} c=\log _{a} c}$

${\log _{b} a^{x} =x\log _{b} a}$

It is essential you know all these basic properties inside out if you want to work with logarithms and exponentials. When the argument of the logarithm is a power of the base you get a simple answer out for the logarithm

${\log _{2} 8} = {\log _{2} 2^{3} =3\log _{2} 2=3}$

${\log _{5} 25} ={\log _{5} 5^{2} =2\log _{5} 5=2}$

We can try to solve some simple equations that involve exponents and logarithms to reinforce our knowledge of the basic rules.

Example 1. Simplify the following expression

$2^{2} 8^{1/4} \left(\cfrac{1}{2} \right)^{.25}$

Solution 1. Everything revolves around 2. Convert to that base each factor. When working with complicated factors write everything upstairs in terms of a positive or negative exponent.

$2^{2} 8^{1/4} 2^{-0.25} =2^{2} 2^{3/4} 2^{-1/4} =2^{2+3/4-1/4} =2^{2+1/2} =4\sqrt{2} =2^{5/2}$

Example 2. Solve the following equation.

${\log _{3} 9^{x^{2} -x} =\log _{3} 27^{x-1} }$

Solution 2. Since the base of the logarithm is three a good starting point is to reexpress the terms of the expression in terms of powers of three.

${\log _{3} 3^{3x^{2} -3x} =\log _{3} 3^{3x-3} }$ ${3x^{2} -3x=3x-3}$ ${x^{2} -2x+1=0}$ ${x=-1}$

Example 3. Solve

$\ln 2x=-\cfrac{1}{2}$

Solution 3. If you have a logarithm alone you can exponentiate both sides of the equation. The natural logarithm has base e so you choose this base.

${\ln 2x=-\cfrac{1}{2} }$ ${2x=e^{-1/2} }$ ${x=\cfrac{1}{2\sqrt{e}}}$

Example 4 Write the following as an exponential with base e.

$2^{x} 3^{-x} e^{x}$

Solution 4. You should know how to hop numbers up into the exponent

$2^{x} 3^{-x} e^{x} =e^{\ln 2^{x} } e^{\ln (1/3)^{x} } e^{x} =e^{(\ln 2)x-(\ln 3)x+x} =e^{(1+\ln 2-\ln 3)x} =e^{(1+\ln (2/3))x}$

Or even simpler you can just write

$\left( \cfrac{2e}{3} \right)^x=e^{x\ln (2e/3)}$

Example 5. Calculate

$x=(\log _{2} 3)(\log _{3} 16)(\log _{4} 125)(\log _{5} 36)(\log _{6} 49)(\log _{7} 8)(\log _{8} 16)$

Solution 5. Since you see a chain of logarithms with the power increasing by one each time you should think we have to use the following formula.

$\log _{a} b\log _{b} c=\log _{a} c$

If we can make a chain all the way from the bases 2 to 8 then all the logarithms will collapse into one. We can just factor numbers and bring their exponents out front.

${x}={(\log _{2} 3)(\log _{3} 16)(\log _{4} 125)(\log _{5}36)(\log _{6} 49)(\log _{7} 8)(\log _{8} 16)}$ $={(\log _{2} 3)(\log_{3} 4^{2} )(\log _{4} 5^{3} )(\log _{5} 6^{2} )(\log _{6} 7^{2} )(\log _{7}8)(\log _{8} 16)}$ $={(2)(3)(2)(2)(\log _{2} 3)(\log _{3} 4)(\log_{4} 5)(\log _{5} 6)(\log _{6} 7)(\log _{7} 8)(\log _{8} 16)}$ $={24\log _{2} 16}$ $={24\log _{2} 2^{4} }=96$

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Exponentials and Logarithms

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