Solution 3.3:6c

From Förberedande kurs i matematik 1

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Before we even start thinking about transforming <math>\log_2</math> and <math>\log_3</math> to ln, we use the log laws
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<center> [[Bild:3_3_6c-1(2).gif]] </center>
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{{NAVCONTENT_STOP}}
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{{Displayed math||<math>\begin{align}
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\log a^b &= b\cdot\log a\,,\\[5pt]
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<center> [[Bild:3_3_6c-2(2).gif]] </center>
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\log (a\cdot b) &= \log a+\log b\,,
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\end{align}</math>}}
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[[Bild:3_3_6_c.gif|center]]
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to simplify the expression
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{{Displayed math||<math>\begin{align}
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\log_{3}\log _{2}3^{118}
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&= \log_{3}(118\cdot\log_{2}3)\\[5pt]
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&= \log_{3}118 + \log_{3}\log_{2}3\,\textrm{.}
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\end{align}</math>}}
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With help of the relation <math>2^{\log_{2}x} = x</math> and <math>3^{\log_{3}x} = x</math> and taking the natural logarithm , we can express <math>\log_{2}</math> and <math>\log_{3}</math> using ln,
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{{Displayed math||<math>\log_{2}x=\frac{\ln x}{\ln 2}\quad</math> and <math>\quad\log_{3}x = \frac{\ln x}{\ln 3}\,\textrm{.}</math>}}
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The two terms <math>\log_3 118</math> and <math>\log_3\log_2 3</math> can therefore be written as
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{{Displayed math||<math>\log_{3}118 = \frac{\ln 118}{\ln 3}\quad</math> and <math>\quad\log_{3}\log_{2}3 = \log_{3}\frac{\ln 3}{\ln 2}\,,</math>}}
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where we can simplify the last expression further with the logarithm law, log (a/b) = log a – log b, and then transform <math>\log _{3}</math> to ln,
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{{Displayed math||<math>\begin{align}
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\log_{3}\frac{\ln 3}{\ln 2}
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&= \log_{3}\ln 3 - \log_{3}\ln 2\\[5pt]
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&= \frac{\ln\ln 3}{\ln 3} - \frac{\ln\ln 2}{\ln 3}\,\textrm{.}
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\end{align}</math>}}
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In all, we thus obtain
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{{Displayed math||<math>\log_{3}\log_{2}3^{118} = \frac{\ln 118}{\ln 3} + \frac{\ln \ln 3}{\ln 3} - \frac{\ln\ln 2}{\ln 3}\,\textrm{.}</math>}}
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Input into the calculator gives
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{{Displayed math||<math>\log_{3}\log_{2}3^{118}\approx 4\textrm{.}762\,\textrm{.}</math>}}
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Note: The button sequence on the calculator will be:
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<center>
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{|
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|1
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|}
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||&nbsp;&nbsp;
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||
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|1
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|}
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||&nbsp;&nbsp;
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|8
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|}
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||&nbsp;&nbsp;
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||
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|LN
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|}
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||&nbsp;&nbsp;
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||
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|÷
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|}
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||
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|3
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|}
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|LN
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|}
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||&nbsp;&nbsp;
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||
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|+
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|}
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|-
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|height="7px"|
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|-
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||
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|3
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|}
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|LN
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|}
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||
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|LN
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|}
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||
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|÷
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|}
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|3
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|}
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||&nbsp;&nbsp;
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|LN
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|}
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|-
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|2
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|-
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|height="7px"|
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|-
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|LN
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|}
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|LN
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|}
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||&nbsp;&nbsp;
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||
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|÷
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|}
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||&nbsp;&nbsp;
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|3
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|}
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||&nbsp;&nbsp;
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|LN
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|}
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||&nbsp;&nbsp;
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{| border="1" cellpadding="3" cellspacing="0"
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|width="30px" align="center"|=
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|}
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</center>

Current revision

Before we even start thinking about transforming \displaystyle \log_2 and \displaystyle \log_3 to ln, we use the log laws

\displaystyle \begin{align}

\log a^b &= b\cdot\log a\,,\\[5pt] \log (a\cdot b) &= \log a+\log b\,, \end{align}

to simplify the expression

\displaystyle \begin{align}

\log_{3}\log _{2}3^{118} &= \log_{3}(118\cdot\log_{2}3)\\[5pt] &= \log_{3}118 + \log_{3}\log_{2}3\,\textrm{.} \end{align}

With help of the relation \displaystyle 2^{\log_{2}x} = x and \displaystyle 3^{\log_{3}x} = x and taking the natural logarithm , we can express \displaystyle \log_{2} and \displaystyle \log_{3} using ln,

\displaystyle \log_{2}x=\frac{\ln x}{\ln 2}\quad and \displaystyle \quad\log_{3}x = \frac{\ln x}{\ln 3}\,\textrm{.}

The two terms \displaystyle \log_3 118 and \displaystyle \log_3\log_2 3 can therefore be written as

\displaystyle \log_{3}118 = \frac{\ln 118}{\ln 3}\quad and \displaystyle \quad\log_{3}\log_{2}3 = \log_{3}\frac{\ln 3}{\ln 2}\,,

where we can simplify the last expression further with the logarithm law, log (a/b) = log a – log b, and then transform \displaystyle \log _{3} to ln,

\displaystyle \begin{align}

\log_{3}\frac{\ln 3}{\ln 2} &= \log_{3}\ln 3 - \log_{3}\ln 2\\[5pt] &= \frac{\ln\ln 3}{\ln 3} - \frac{\ln\ln 2}{\ln 3}\,\textrm{.} \end{align}

In all, we thus obtain

\displaystyle \log_{3}\log_{2}3^{118} = \frac{\ln 118}{\ln 3} + \frac{\ln \ln 3}{\ln 3} - \frac{\ln\ln 2}{\ln 3}\,\textrm{.}

Input into the calculator gives

\displaystyle \log_{3}\log_{2}3^{118}\approx 4\textrm{.}762\,\textrm{.}


Note: The button sequence on the calculator will be:


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