Question 167703
<font size = 7 color = "red"><b>Edwin's solution follows:</b></font>

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? 
<pre><font size = 4 color = "indigo"><b>
You could get your calculator and find {{{log(1776)}}} which is
{{{3.249442961}}} and then mutiply that by {{{53}}} to get
{{{172.220477}}}. However that won't get you any closer
to finding {{{1776^53}}}!

That's because calculators as a rule can only work 
directly with numbers less than {{{10^100}}}, "1 google".

You can't get {{{1776^53}}} 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 {{{1776^30}}} 
directly as {{{matrix(1,3,3.042908149, "×", 10^97)}}}

However when I try to get {{{1776^31}}} I get an overflow
error message.  

We can break {{{1776^53}}} down this way:

{{{1776^53=(1776^30)(1776^23)}}} both of which can be 
found on the calculator:

{{{matrix(1,5,1776^30, "=", 3.042908149, "×", 10^97)}}}  
{{{matrix(1,5,1776^23, "=", 5.459943055, "×", 10^74)}}}

Then we can multiply those decimal parts, {{{matrix(1,5,3.042908149, "×",5.459943055,"=",16.61410521)  }}} 

then add the exponents of ten and get {{{matrix(1,5,  10^97, "×", 10^74, "=", 10^171)}}}

So therefore

{{{matrix(1,5, 1776^53, "=", 16.61410521, "×", 10^171)}}}

However, that's not in scientific notation because {{{16.61410521}}}
is not less than 10, so we write it in scientific notation as
{{{matrix(1,3 ,1.661410521, "×", 10^1)}}}, then substituting we have

{{{matrix(1,7, 1776^53, "=", 1.661410521, "×", 10^1, "×", 10^171)}}}

or adding the exponents of ten:

{{{matrix(1,5, 1776^53, "=", 1.661410521, "×", 10^172)}}}

Notice that we could have broken {{{1776^53}}} down in a number
of other ways, for example:

{{{matrix(1,13, 1776^53, "=", 1776^26, "×", 1776^27, "=", 3.058559951, "×", 10^84, "×", 5.43200247, "×", 10^87)}}}

so
{{{matrix(1,13, 1776^53, "=", 3.058559951, "×", 10^84, "×", 5.43200247, "×", 10^87,"=", 16.61410499,"×", 10^171 )}}}.

And so we have, as above:

{{{matrix(1,5, 1776^53, "=", 1.661410499, "×", 10^172)}}}

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: 

{{{matrix(1,5, 1776^53, "=", 1.6614105, "×", 10^172)}}}

Edwin</pre>