+0

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

0
298
2
+4

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

maria3080j  Apr 6, 2015

#1
+84159
+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
Sort:

#1
+84159
+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
#2
+91928
+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

35 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details