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a=log(sqrt(10))

b=log(sqrt(10)+1)

c=log(sqrt(10)+2)

then

b^2=?

ac=?

Guest Jun 8, 2014

Best Answer 

 #4
avatar+26399 
+5

I did use brackets - Math (Input=Result) removed them!  Try it!

Alan  Jun 9, 2014
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5+0 Answers

 #1
avatar+91435 
+3

a=log(sqrt(10)) = 1/2

b=log(sqrt(10)+1) = 

c=log(sqrt(10)+2)

then

    $$WRONG\qquad b^2=log(\sqrt{10}+1)^2=2log(\sqrt{10}+1)\qquad WRONG$$      

As Alan has pointed out -

this previous answer is incorrect, question is not of the from log(x2) it is (log x)2 so it should be 

 

$$b^2=[log(\sqrt{10}+1)]^2 \approx 0.38357$$ 

 

$$ac=\frac{log(\sqrt{10}+2)}{2}$$

Melody  Jun 8, 2014
 #2
avatar+26399 
+5

Careful with b2

$${\mathtt{b}} = {log}_{10}\left({\sqrt{{\mathtt{10}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right) \Rightarrow {\mathtt{b}} = {\mathtt{0.619\: \!331\: \!048\: \!066\: \!094\: \!5}}$$

$${\mathtt{bsquared}} = {{log}_{10}\left({\sqrt{{\mathtt{10}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}^{\,{\mathtt{2}}} \Rightarrow {\mathtt{bsquared}} = {\mathtt{0.383\: \!570\: \!947\: \!098\: \!647\: \!1}}$$

$${\mathtt{2}}{\mathtt{\,\times\,}}{log}_{10}\left({\sqrt{{\mathtt{10}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right) = {\mathtt{1.238\: \!662\: \!096\: \!132\: \!189}}$$

Alan  Jun 8, 2014
 #3
avatar+91435 
0

Thank you Alan,  i appreciate you pointing out this error.  

I am surprised that you didn't use brackets in your answer because without brackets the meaning is not clear.

Melody  Jun 9, 2014
 #4
avatar+26399 
+5
Best Answer

I did use brackets - Math (Input=Result) removed them!  Try it!

Alan  Jun 9, 2014
 #5
avatar+91435 
0

 

(log(sqrt(10)+1))^2=

$${{log}_{10}\left({\sqrt{{\mathtt{10}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}^{\,{\mathtt{2}}} = {\mathtt{0.383\: \!570\: \!947\: \!098\: \!647\: \!1}}$$

So it did - another wrinkle for Andre to iron out!

Sorry Alan.

 

PS I have already sent Andre an email telling him of this probem.  I am assuming that he puts calculator errors near the top of his priority list!

Melody  Jun 9, 2014

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