Question 167703This question is from textbook Algebra and Trigonometry
: In general, how do I find the logarithm of the answer
ex: evaluate 1776^53
I can get 53 Log 1776 = log x, but what is next?
This question is from textbook Algebra and Trigonometry
Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1776^53
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Let x = 1776^53
log(x)= 53*log(1776)
log x = 53*3.24944...
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10^(logx) = 10^(53*3.24944...)
x = 10^172.2203...
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Cheers,
Stan H.
Answer by Edwin McCravy(20060)  (Show Source):
You can put this solution on YOUR website! Edwin's solution follows:
In general, how do I find the logarithm of the answer
ex: evaluate 
I can get  , but what is next?
You could get your calculator and find  which is
 and then mutiply that by  to get
 . However that won't get you any closer
to finding !
That's because calculators as a rule can only work
directly with numbers less than  , "1 google".
You can't get on a calculator directly because
that number is bigger than googol. Most calculators made today
can do anything but give an overflow error message for a google
and anything larger.
However you can get larger numbers indirectly on your calculator.
On my calculator I find that I can get 
directly as 
However when I try to get  I get an overflow
error message.
We can break down this way:
 both of which can be
found on the calculator:
Then we can multiply those decimal parts, 
then add the exponents of ten and get 
So therefore
However, that's not in scientific notation because 
is not less than 10, so we write it in scientific notation as
 , then substituting we have
or adding the exponents of ten:
Notice that we could have broken down in a number
of other ways, for example:
so
 .
And so we have, as above:
There is a slight difference between the two. I would guess
that that latter is closer to the true value, but we could
safely round either answer to this:
Edwin
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