Question

write in logarithmic form:-E=1/2*mv^2

Answer 1

E = 1/2 mv²  

Log E = Log(1/2 mv²) = Log1/2 + Log m + Log v²
Log E = Log (1/2) + Log m + 2Log v
          =  Log 1 - Log 2 + Log m + 2Log v = 0 - 0.301 + Logm + 2log v
 
Hope this assists.

Answer 2

The logarithm form is :

$\displaystyle \sf{log \:E = log \: m + 2 \: log \: v - log \: 2 }$

Given :

$\displaystyle \sf{ }E = \frac{1}{2} m {v}^{2}$

To find :

The logarithm form

Formula :

We are aware of the formula on logarithm that

$\sf{1. \: \: \: log( {a}^{n} ) = n log(a) }$

$\sf{2. \: \: log(ab) = log(a) + log(b) }$

$\displaystyle \sf{3. \: \: log \bigg( \frac{a}{b} \bigg) = log(a) - log(b) }$

$\sf{4. \: \: log_{a}(a) = 1}$

Solution :

Step 1 of 2 :

Write down the given expression

Here the given expression is

$\displaystyle \sf{ }E = \frac{1}{2} m {v}^{2}$

Step 2 of 2 :

Express in logarithm form

$\displaystyle \sf{ }E = \frac{1}{2} m {v}^{2}$

Taking logarithm in both sides we get

$\displaystyle \sf{ }log \: E = log \: \bigg[ \frac{1}{2} m {v}^{2}\bigg]$

$\displaystyle \sf{ \implies }log \: E = log \: \bigg(\frac{1}{2} \bigg) + log \:m + log \:( {v}^{2})\: \: \: \bigg[ \: \because \:log(ab) = log(a) + log(b) \bigg]$

$\displaystyle \sf{ \implies }log \: E = \bigg(log \: 1 -log \: 2 \bigg) + log \:m + 2log \:v\: \: \: \bigg[ \: \because \:log( {a}^{n} ) = n log(a) \bigg]$

$\displaystyle \sf{ \implies }log \: E = -log \: 2 + log \:m + 2log \:v$

$\displaystyle \sf{ \implies }log \: E = log \:m + 2log \:v-log \: 2$

Which is the required logarithm form of the expression

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