First approximation
x = 5^123
log(5) = 0.69897000433
log(x) = 123 log(5) = 85.97331053259
x = 10^85 times antilog(0.97331053259)
x = 9.4039548 x 10^85
----
125*5^120
125*(5^30)^4
125*(25^15)^4
125*[(25^5)^3]^4
25^5 = 9,765,625
(25^5)^3 = 95,367,441,640,625 * 9,765,625
(25^5)^3 = 931,322,673,171,728,515,625
square this twice, then multiply by 125.
2007-10-11 15:59:09
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answer #1
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answered by Raymond 7
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MS Excel solves this very easily. The answer I've got is 9,40395E+85
2007-10-11 15:59:48
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answer #2
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answered by Joecat73 3
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~9.4039548065783000 E 273
2007-10-11 15:56:54
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answer #3
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answered by gruntparty 2
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9.403954806578300063749892297778e+85
in notation form
2007-10-11 15:59:07
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answer #4
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answered by rahilaw 2
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94,039,548,065,783,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000
2007-10-11 15:54:54
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answer #5
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answered by Anonymous
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