Asked by MAths lover Please Helppppp
We are given that log102<0.302. How many digits are there in the decimal representation of 5^500?
Answers
Answered by
Reiny
I am sure you meant:
log<sub>10</sub> 2 < .302
(you could have just said log 2 ≤ .302 , if the base is omitted it is assumed to be base 10)
let x = 5^500
= (10/2)^500
log x = log (10/2)^500
log x = 500(log (10/2)
= 500(log10 - log2)
= 500(1 - log2)
= 500 - 500log2
= 500 - 151 , = 349 from your given
= 349.48 by using calculator for log2
so your log x > 349
what does that mean?
if logx = 2 , then x = 10^2 ---100 --- 3 digits
if logx = 3, then x = 10^3 -- 1000 --- 4 digits
...
if logx = 349 , then x = 10^349 ----- 350 digits
but log x > 349 , so it must contain 351 digits
log<sub>10</sub> 2 < .302
(you could have just said log 2 ≤ .302 , if the base is omitted it is assumed to be base 10)
let x = 5^500
= (10/2)^500
log x = log (10/2)^500
log x = 500(log (10/2)
= 500(log10 - log2)
= 500(1 - log2)
= 500 - 500log2
= 500 - 151 , = 349 from your given
= 349.48 by using calculator for log2
so your log x > 349
what does that mean?
if logx = 2 , then x = 10^2 ---100 --- 3 digits
if logx = 3, then x = 10^3 -- 1000 --- 4 digits
...
if logx = 349 , then x = 10^349 ----- 350 digits
but log x > 349 , so it must contain 351 digits
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