#3**+10 **

Although this calculator says "infinity", you can get a better answer by using logs.

Let x = 7^{2015}

Then log(x) = log(7^{2015})

---> log(x) = 2015·log(7)

---> log(x) = 1702.87 (approx)

---> x = 10^{1702.87}

Since 10^{.87} is approximately 7,

the value is a 7 followed by 1702 zeros (approximately).

geno3141
Mar 2, 2015

#3**+10 **

Best Answer

Although this calculator says "infinity", you can get a better answer by using logs.

Let x = 7^{2015}

Then log(x) = log(7^{2015})

---> log(x) = 2015·log(7)

---> log(x) = 1702.87 (approx)

---> x = 10^{1702.87}

Since 10^{.87} is approximately 7,

the value is a 7 followed by 1702 zeros (approximately).

geno3141
Mar 2, 2015