Solution 3.3:6c

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Before we even start thinking about transforming
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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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<math>\log _{2}</math>
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and
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<math>\log _{3}</math>
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to ln, we use the log laws
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<math>\lg a^{b}=b\centerdot \lg a</math>
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<math>\lg \left( a\centerdot b \right)=\lg a+\lg b</math>
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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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\log (a\cdot b) &= \log a+\log b\,,
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\end{align}</math>}}
to simplify the expression
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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<math>\begin{align}
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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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& \log _{3}\log _{2}3^{118}=\log _{3}\left( 118\centerdot \log _{2}3 \right) \\
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& =\log _{3}118+\log _{3}\log _{2}3 \\
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\end{align}</math>
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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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With help of the relation
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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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<math>2^{\log _{2}x}=x</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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and
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<math>3^{\log _{3}x}=x</math>
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-
+
-
 
+
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and taking the natural logarithm , we can express
+
-
<math>\log _{2}</math>
+
-
and
+
-
<math>\log _{3}</math>
+
-
using ln,
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-
 
+
-
 
+
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<math>\log _{2}x=\frac{\ln x}{\ln 2}</math>
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-
and
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<math>\log _{3}x=\frac{\ln x}{\ln 3}</math>
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The two terms log3 118 and log3 log2 3 can therefore be written as
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<math>\log _{3}118=\frac{\ln 118}{\ln 3}</math>
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and
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<math>\log _{3}\log _{2}3=\log _{3}\left( \frac{\ln 3}{\ln 2} \right),</math>
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+
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+
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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
+
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<math>\log _{3}</math>
+
-
to ln,
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-
 
+
-
 
+
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<math>\begin{align}
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& \log _{3}\left( \frac{\ln 3}{\ln 2} \right)=\log _{3}\ln 3-\log _{3}\ln 2 \\
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& =\frac{\ln \ln 3}{\ln 3}-\frac{\ln 2}{\ln 3} \\
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\end{align}</math>
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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>}}
In all, we thus obtain
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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<math>\log _{3}\log _{2}3^{118}=\frac{\ln 118}{\ln 3}+\frac{\ln \ln 3}{\ln 3}-\frac{\ln 2}{\ln 3}</math>
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Input into the calculator gives
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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<math>\log _{3}\log _{2}3^{118}\approx 4.762</math>
 
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NOTE: the button sequence on a calculator will be:
 
 +
Note: The button sequence on the calculator will be:
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<math>\begin{align}
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<center>
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& \left[ 1 \right]\quad \left[ 1 \right]\quad \left[ 8 \right]\quad \left[ \text{LN} \right]\quad \left[ \div \right]\quad \left[ 3 \right]\quad \left[ \text{LN} \right]\quad \left[ + \right] \\
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{|
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& \left[ 3 \right]\quad \left[ \text{LN} \right]\quad \left[ \text{LN} \right]\quad \left[ \div \right]\quad \left[ 3 \right]\quad \left[ \text{LN} \right]\quad \left[ - \right]\quad \left[ 2 \right] \\
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||
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& \left[ \text{LN} \right]\quad \left[ \text{LN} \right]\quad \left[ \div \right]\quad \left[ 3 \right]\quad \left[ \text{LN} \right]\quad \left[ = \right] \\
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{| border="1" cellpadding="3" cellspacing="0"
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& \quad \\
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|width="30px" align="center"|1
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\end{align}</math>
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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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||
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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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||&nbsp;&nbsp;
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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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||
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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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||&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"|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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||
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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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||
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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"|2
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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"|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"|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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||&nbsp;&nbsp;
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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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||
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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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</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:


1
  
1
  
8
  
LN
  
÷
  
3
  
LN
  
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3
  
LN
  
LN
  
÷
  
3
  
LN
  
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2
LN
  
LN
  
÷
  
3
  
LN
  
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