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3^352^log(10^(sqrt(sqrt3(

maria3080j  Apr 6, 2015

Best Answer 

 #1
avatar+80865 
+10

I'm assuming this is

3^352^(log10^(3^(1/4))

3^(1/4) = 1.3160740129524925  ... so we have

3^352^log10^1.3160740129524925 =

3^352^1.3160740129524925 =

3^463.25805255927736

In terms of base 10 .... 3 = 0.4771212547196624

So, we have

(10^0.4771212547196624)^463.25805255927736  =

10^221.030263296069725261145127163264 =

10^221 * 10^0.030263296069725261145127163264 = 

10^221 * 1.072   {approximately } 

Or, in scientifiic notation

1.072 x  10^221  = 1072 followed by 218 zeroes..... (more or less)

 

 

  

CPhill  Apr 6, 2015
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2+0 Answers

 #1
avatar+80865 
+10
Best Answer

I'm assuming this is

3^352^(log10^(3^(1/4))

3^(1/4) = 1.3160740129524925  ... so we have

3^352^log10^1.3160740129524925 =

3^352^1.3160740129524925 =

3^463.25805255927736

In terms of base 10 .... 3 = 0.4771212547196624

So, we have

(10^0.4771212547196624)^463.25805255927736  =

10^221.030263296069725261145127163264 =

10^221 * 10^0.030263296069725261145127163264 = 

10^221 * 1.072   {approximately } 

Or, in scientifiic notation

1.072 x  10^221  = 1072 followed by 218 zeroes..... (more or less)

 

 

  

CPhill  Apr 6, 2015
 #2
avatar+91432 
+5

3^352^log(10^(sqrt(sqrt3(

$$\\y=3^{{352}^{log\left(10^\sqrt{\sqrt{3}}}\right)}}\\\\
y=3^{{352}^{log\left(10^{3^{0.25}}}}\right)}}\\\\
y=3^{{352}^{\left(3^{0.25}log10}\right)}}\\\\
y=3^{{352}^{\left(3^{0.25}}\right)}}\\\\
log(y)=log\left(3^{{352}^{\left(3^{0.25}}\right)}}\right)\\\\
log(y)={352}^{(3^{0.25})}log(3)\\\\
log(y)=1071.685578\\\\
y\approx 10^{1071.7}\\\\$$

 

This answer is different from Chris's and these large powers really confuse me.

Could other mathematicians (including Chris) take a look and comment please.  

-----------------------------------------

Melody  Apr 6, 2015

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